polyominoes2.2 - code-spot

H E R M A N T U L L E K E N

P O LYO M I N O E S 2 . 2

H O W T H E Y F I T T O G E T H E R .

F I R S T P U B L I S H E D : 1 M AY 2 0 1 8L A S T M O D I F I E D : 1 9 M AY 2 0 1 9

Contents

1 Preface 9

1.1 Acknowledgments 10

1.2 Notes 10

2 Introduction 11

2.1 A Quick Introduction to Polyominoes and Tilings 11

2.2 Symmetry 14

2.3 The Geometry and Topology of a Polyomino 16

2.4 Further Reading 23

2.5 Hexominoes, and Heptominoes 25

3 Dominoes I 27

3.1 Tiling Criteria 28

3.1.1 Flow 30

3.1.2 The Marriage Theorem 39

3.1.3 Cylinder Deletion 42

3.1.4 Coloring 50

3.2 The Structure of Domino Tilings 53

3.2.1 Closed strips 53

3.2.2 Frozen Cells 57

3.2.3 Tiling Transformations 60


3

4 Dominoes II 70

4.1 The Structure of Domino Tilings 70

4.1.1 Height Functions 70

4.1.2 Forced flippable pairs 87

4.2 Counting Tilings 96

4.2.1 Rectangles 96

4.2.2 Aztec Diamond 100

4.2.3 Arctic Behavior 103

4.3 Constraints and Monominoes 103

4.3.1 Optimal Tilings 103

4.3.2 Tatami Tilings 111

4.3.3 Fault-free tilings 111

4.3.4 Even Polyominoes 115

4.4 Summary of tiling criteria 119

4.5 Miscellaneous 119

4.5.1 Polyominoes that can be Tiled by Dominoes 119


5 Various Topics 124

5.1 The Tiling Hierarchy 124

5.1.1 Strips, Crossed Strips, Bent Strips, and Half Strips 129

5.1.2 Extending the hierarchy 131

5.1.3 Further Reading 132

5.2 Colorings 132

5.3 Border Words 151

6 Rectangles 153

6.1 Tilings by Rectangles 154

6.1.1 Which rectangles can be tiled by a set of rectangles? 154

6.1.2 The gap number of rectangular tilings 163

6.1.3 Tiling by Rectangles of other Regions 165

4

6.2 Order 165

6.2.1 Order Theorems 166

6.2.2 L-shaped polyominoes 173

6.2.3 Known Orders 175

6.2.4 The orders of small polyominoes 185

6.3 Prime Rectangles 185

6.3.1 Trominoes 189

6.3.2 Tetrominoes 190

6.3.3 Pentominoes 193

6.3.4 Hexominoes 195

6.3.5 Other polyominoes 198

6.3.6 Prime Rectangles of Small Polyominoes 199

6.4 Fault-free tilings 201

6.4.1 Rectangles 201

6.4.2 Trominoes 205

6.4.3 Tetrominoes 207

6.4.4 Pentominoes 209

6.4.5 Extensions 218

6.5 Simple Tilings 218


7 Plane Tilings 225

7.1 Tilings by Translation 226

7.2 Tilings by Translation and Rotation 226

7.3 Isohedral and k-Isohedral Tilings 227

7.4 m-morphic and p-poic Polyominoes 237

7.5 Non Tilers 241

7.6 Aperiodic Tilings 242


8 Various Topics II 251

8.1 Compatibility 251

8.2 Exclusion 253

8.3 Fountain sets 255

5

Bibliography 256

Index 264

List of Tables

1 Notation 10

2 Names for classes of polyominoes 11

3 The symmetry types of free polyominoes. 15

4 The number of maximally biased polyominoes. 33

5 Maximum deficiency. 33

6 Number of free snakes S(k) compared to the number if free poly-ominoes P(k). 57

7 ρ(1) for R(m, n). 84

8 ρ(1) for A(n). 84

9 Number of domino tilings of a m× n rectangle. 100

10 Number of tilings of the Aztec diamond. 102

11 Gap numbers for some classes of polyominoes. 107

12 Number of tilings of odd-area rectangles with one monomino anddominoes. 109

13 Summary of tiling criteria for domino tilings. 120

14 Balanced, tileable and uniquely tileable dominoes. 121

15 Tiling Classes. For example, if some polyomino P ∈ R it means itcan tile a rectangle. 125

16 Tiling classes for polyominoes. The table gives the smallest class apolyomino is part of. Classes marked with an asterisk is not guar-anteed to be the smallest. An accent shows the polyomino is not inthat class. 128

17 Prime Half Strips and Strips. 130

18 Computed values of C(m, n). 135

19 Colorings that realize the number of colors in Table 18. 136

20 Chromatic numbers for small polyominoes. Intervals [a, b] denotelower and upper bounds. Numbers in brackets show the upper boundis provided by an explicit coloring rather than a theorem. 144

21 Chromatic numbers for small polyominoes. 145


7





27 Border words for tetrominoes. 151

28 Characterization of tilings of rectangles by rectangles. 162

29 Odd orders for various L-polyominoes. 174

30 Examples of orders. 175

31 Examples of odd orders. 184

32 Table giving order information about polyominoes. 188

33 Summary of proofs for pentominoes that do not tile rectangles. 195

34 Prime Rectangles of polyominoes 200

35 Fault-free Rectangles of polyominoes 201

36 BN factorizations for various polyominoes and families of polyomi-noes. 227

37 The 9 ways a tile can tile the plane. 233

38 Table showing counts for the number of various polyominoes thattile a certain way. From Myers. 234

39 Table showing number of polyominoes that are k-anisohedral My-ers. 235

40 Polyominoes that tile the plane by translation Myers 236

41 Examples of m-morphic and p-poic polyominoes. 238

42 The number of polyominoes that are m-morphic Myers. (Meyers orig-inally classified the monomino as monomorphic—presumably be-cause he allows non-discrete tilings.) 240

43 Heesch Numbers for Polyominoes Kaplan (2017) 242

44 Table showing the least number the number of tiles necessary to tilethe overall LCM (as far as is known) 251

45 Table showing the least number the number of tiles necessary to tilethe hole-less LCM (as far as is known) 252

46 Inverse frequency for rectangles. (Adapted from Klamkin and Liu(1980)) 253

1Preface

On 8 September 2017 I set out to discover as much about polyomi-noes as I could. I thought it would be nice to collect what I found ina few essays, and this collection is the result.

In putting these essays together, I tried to keep the mathematicssimple; you will not need to know much graph theory, combinatorics,or other machinery to understand them. At times this means I coveronly a very concrete special case of a more general (and arguablybeautiful) result, and some proofs in their naive disguise may be a bitclunky; this is the trade-off. I do try to give as many references to thebigger theory as possible.

These essays are work in progress. This means I have not yetcaught every error or included every reference, and I omitted orneglected (for now) some important topics. If you spot an error orhave any suggestions, please let me know 1. 1 [email protected]

Currently, I am fleshing out some of the newer sections. But I hopeto add more to this; some topics I am considering are Enumeration,Algorithms, Compatibility, and Tatami Tilings.

Problems. There are three types of problems:

• Exercises that are interesting ideas or problems that occurredto me as I worked through the topics that I thought is helpfulin building intuition and understanding, but not importantenough for the main text.

• Questions that occurred to me that I have not solved. Some-times these are ideas for further investigation on which I didnot spend any time; in other cases these are problems I couldnot solve even after some effort.

• Open questions from the literature that usually has been openfor some time; most of these can be considered very difficult.

mailto:[email protected]

10 polyominoes

I plan to mark these in future version, but for now these are all just“Problems”.

1.1 Acknowledgments

I want to thank Justin Southey for reading a draft of the first essay.He made many suggestions that improved it greatly.

1.2 Notes

Numbers in orange like A000105 refer to integer sequences in theOnline Encyclopedia of Integer Sequences (Sloane). Clicking on themtakes you to the relevant entry.

Notation Meaning

|R| The area of R, the number of cells of R∆(R) Deficiency of Rφ(R) The flow of RW(R),B(R) The set of white (black) cells of Rw(S), b(S) The number of dominoes with one white (black) cell inside S and the other outsideGT (R), G(R) The gap number of region R with respect to tileset Tρ(T) The ranking of a tilingµ(T, U) The number of moves from tiling T to U#T R, #R The number of tilings of a region R by tiles from the T Table 1: Notation

https://oeis.org/A000105

2Introduction

n Name

1 Monomino2 Domino3 Tromino4 Tetronino5 Pentomino

Table 2: Names for classes of polyomi-noes

2.1 A Quick Introduction to Polyominoes and Tilings

Polyominoes1

are shapes formed by “gluing” 1× 1 squares to- 1 A polyomino with n cells is alsocalled an n-omino.gether, edge-to-edge. Figures 1 to 7 show the polyominoes with five

cells or less. The squares that make up a polyomino are called cells,and we classify polyominoes according to the number of cells theyhave. The names for the smaller classes of polyominoes is given inTable 2. We will use the notation Pfor the set of all polyominoes, andP\for the set of polyominoes with n cells.

Figure 1: Monomino

Figure 2: Domino

Figure 3: Bar tromino

Figure 4: Right tromino

These simple shapes give rise to a variety of interesting prob-lems. The one which will be the main topic of these essays, is: Whichshapes can we build if we put them together so that edges line up?

Problems such as these are called tiling problems. A different wayto put a tiling problem is: given some shape (we call this a region2)

2 Also called figure or picture.

and a set of other shapes (called the tiles), can we cover the regioncompletely with copies of the tiles so that no piece of tile falls outsidethe region, no tiles overlap? If this is possible, we say the region canbe tiled by the set of tiles. Figure 5 shows some examples of tilings.

Tiling is a big and difficult topic in mathematics, but we can getsome insight by limiting our study to polyominoes.

In our case, we will only work with regions that are also composedfrom squares glued edge-to-edge. Regions are often polyominoesthemselves, but we will at times also consider infinite regions, or dis-connected regions, or regions with barriers that tiles are not allowedto cross.

The number of cells of a finite region R is denoted by |R|. Sinceeach cell is has an area of 1, |R| also equals the area.

In a region or a polyomino, we call two cells neighbors if theyshare an edge. A cell with four neighbors is called an interior cell;otherwise it is called a border cell. A cell with only one or twoneighbors is called a corner cell.

12 polyominoes

(a) A tiling of a rectangle by the Ttetromino.

(b) A tiling of a rectangle by Y pentominoes.

(c) A tiling of a rectangle by a heptomino.

Figure 5: Examples of tilings.

(a) Square (b) T (c) Skew (d) L (e) BarFigure 6: Tetrominoes

introduction 13

(a) F (b) I (c) L (d) N (e) P (f) T

(g) U (h) V (i) W (j) X (k) Y (l) ZFigure 7: Pentominoes

Two cells are connected if we can move from neighbor to neigh-bor from one to the other. We call a region where every cell is con-nected to every other cell connected, and if it has no holes3, we call 3 Although it is intuitively clear what

a hole is, it is a bit harder to give aprecise definition, and we will not dothis here.

it simply-connected.4 Figure 8 shows examples of polyominoes that

4 Polyominoes with holes are also calledholey; polyominoes without holes arealso called simple (Herzog et al., 2015)or profane (Toth et al., 2017, p. 367).Some authors use a slightly differentnotion of simply-connected, so that thesecond polyomino in Figure 8 is simplyconnected.

are not simply-connected as we use the term here.

Figure 8: Two examples of polyominoeswith holes.

To conclude this quick introduction, I state two obvious theoremsthat we will reference often. I omit their proofs.

Theorem 1 (Area Criterion). If we have a tile set where all tiles have arean, then we can only tile regions whose area is divisible by n.

[Referenced on pages 13, 27, 30, 50, 90, 101, 106, 108, 119, 155, 166,194, 197 and 210]

Example 1. If a region is tileable by dominoes, then its area is even.

If S is a subset of the cells of a region R, we say S is a subregion ofR, (see Figure 9), and if in addition S 6= R, S is a proper subregionof R. The set of cells in R not in S is denoted R− S. A partition of aregion R is a collection of subregions Si of R such that every cell of Ris contained in exactly one of Si . To partition a region means to givea partition of the region.

Theorem 2 (Partitions). If all the subregions in a partition of a region aretileable, then so is the region.

[Referenced on pages 49, 54, 56, 58, 59, 117 and 118]

Figure 9: In this example, the 4 × 4square is the region, and both theyellow and blue regions are subregionsof the square.

Example 2. There is a region with 2n cells not tileable by P\for all n.Consider a cross with area 2n and arms of length bn/2c or bn/2c+ 1. If

we place a polyomino to cover one arm, it must cover the center of the cross,and another arm. This partitions the cross into three pieces—the covered

14 polyominoes

arms and the two non-covered arms. Each of the non-covered arms is smallerthan n and non-zero, so by Theorem 1 cannot be tiled. And hence, the wholefigure cannot be tiled.

2.2 Symmetry

Generally, we allow polyominoes to be rotated or reflected any waypossible for tiling problems, and therefor we do not distinguish be-tween different orientations or reflections of a polyomino. In thiscase, we call the polyominoes free (Redelmeier, 1981, Section 3). Ifwe distinguish reflections, but not rotations, we call the polyominoesone-sided (Golomb, 1996, p. 70). If we distinguish between both re-flections and rotations, we call the polyominoes fixed (Redelmeier,1981, Section 3). With this terminology, there are 7 one-sided tetromi-noes, and 19 fixed tetrominoes.

Figure 10: One-sided tetrominoes.

Figure 11: The set of fixed tetrominoes.

If a transformation leaves a free polyomino unaffected, we say thepolyomino has the symmetry associated with that transformation.For example, the skew polyomino is not affected by 180° rotations, soit has 180°-rotational symmetry.

The symmetry index of a (free) polyomino is the number of fixedpolyominoes congruent to it (Redelmeier, 1981, Section 3).

introduction 15

Table 3 shows details of the symmetry types, including an exampleof each.

Type Index Symmetries Smallest Example

None 8

Rot 4 180° rotation

Axis 4 horizontal or vertical reflection

Diag 4 diagonal-reflection

Rot2 2 90° rotation

Axis2 2 horizontal and vertical reflection or 180° rotation

Diag2 2 diagonal-reflection or 180° rotation

All 1 all

Table 3: The symmetry types of freepolyominoes.

Tilings can also be classified by their symmetries in a similar way.A tiling of a region R cannot have more symmetries than R.

Theorem 3. If a region has a tiling by a polyomino with symmetry typeAll, the tiling is unique.

[Referenced on pages 166 and 201]

Proof. (Adapted from nickgard (2017).) The left-most cell in the toprow of the region can only be tiled by the left-most cell of the top rowof the tile in any orientation (which are all equivalent because the tilehas symmetry type All). We can form a new region from the untiledpart, which by a similar argument can only be tiled one way. We canrepeat this until eventually the complete figure is tiled. Since in eachstep only one placement is possible, the entire tiling is unique.

Since squares have symmetry type All, it follows that if a regionhas a tiling by a single square, the tiling is unique.

Problem 1. Show that for a polyomino P ∈ All with perimeter p,

(1) |P| ≡ 0 or 1 (mod 4)

(2) p(P) ≡ 0 (mod 4).

Problem 2. A Baiocchi figure of a polyomino is a region with symmetrytype All that is tileable by that polyomino5. A Baiocchi figure of a polyomino 5 The idea was suggested by Claudio

Baiocchi in January 2008, and appearedin that month in Friedman (2008).

16 polyominoes

is minimal if there is no smaller Baiocchi figure for that polyomino. Figure12 shows minimal Baiocchi figures for the tetrominoes, and Figure 13 showsholeless variants.

Figure 12: Minimal Baiocchi figures forthe tetrominoes (Sicherman, 2015).

Figure 13: Holeless variants for minimalBaiocchi figures for the L-tetromino andskew tetromino (Sicherman, 2015).

(1) Find minimal Baiocchi figures for the pentominoes.

(2) Is there a Baiocchi figure for the U-pentomino without holes? Thisseems to be an open problem (Sicherman, 2015).

(3) Find the minimal holeless Baiochhi figures for other pentominoes.

2.3 The Geometry and Topology of a Polyomino

(Note: this section needs to be reworked completely; as it stands ithas a lot of problems, especially with vague definitions that make theproofs unconvincing. I left this section in, because some of the ideashere are relevant to other sections.)

A rectilinear polygon6 is a polygon with all edges meeting at right

6 Also called orthogonal polygon oraxis-aligned.

angles. A polyomino, therefor, is a rectilinear polygon whose edgesare integer lengths. We prove a few simple facts about rectilinearpolygons that will be useful to our study of polyominoes.

Theorem 4 (Wikipedia contributors (2017)). In a simply-connectedrectilinear polygon, the number of vertical and horizontal edges are equal.


Proof. Each vertical edge is followed by a horizontal edge, and viceversa.

Theorem 5. In a polyomino the total horizontal edge length and totalvertical edge length are even, and so is the perimeter.

[Referenced on pages 16, 21 and 37]

Proof. If traveling along the perimeter we go x units left, and x′ unitsright, then x + x′ = 0 (because the perimeter is a closed finite curve,so we eventually end up where we started). Then the total horizontaldistance is |x|+ |x′| = 2|x|, which is even since x is an integer. Thesame goes for the vertical length.

Since the perimeter is the sum of the vertical and horizontallength, which are both even, it must be even too.

Theorem 6 (Csizmadia et al. (2004), Lemma 3.3). If the lengths of eitherall horizontal or all vertical edges are odd, then there is an even number ofhorizontal and vertical edges, and the total number of edges is divisible byfour.

introduction 17

[Referenced on page 37]

Proof. WLG suppose the lengths of all horizontal sides are odd. Sincethe total length must be even (Theorem 5), there must be an evennumber of them, and because the number of horizontal and verticalsides are equal (Theorem 4), the total must be divisible by four.

Problem 3 (Sallows et al. (1991)). A golygon7 is a rectiliear polygon 7 Also called a 90°serial isogon.

whose edge lengths forms a sequence 1, 2, 3, 4, · · · ; usually, self-intersectionsare allowed. An example with 8 sides is shown in Figure 14. Show thenumber of edges of an golygon must be divisible by 8.

Figure 14: Golygon with 8 sides. Thisisogon tiles the plane.

The golygon in Figure 14 is the only one with 8 sides. It is the onlyknown golygon that can tile the plane. It seems to be an open problemwhether any other golygons can tile the plane (Sallows, 1992).

Theorem 7. In a simply-connected rectilinear polygon, the number of edgesis even.


Proof. Since there is the same number of vertical edges as horizontaledges (Theorem 4), their sum is even.

A corner with interior angle of 90° is called convex and a cornerwith an interior angle of 270° is called concave Bar-Yehuda and Ben-Hanoch (1996).

Theorem 8 (Wikipedia contributors (2017)). In a finite, simply-connectedrectilinear polygon, the number of convex corners is 4 more than the numberof concave corners.


Proof 1. Suppose we travel anti-clockwise around the perimeter.At concave corners, we need to make a left turn, and at a convexcorner, we need to make a right turn. When we return to our startingpoint, we face the original direction, after having made a net turn of360 degrees, which means we had to make at least 4 right turns, ortraversed 4 convex corners.

Proof 2. The sum of interior angles of a polygon with k sides is180◦(k − 2). Suppose we have m convex corners and n convex cor-ners. We then have the equation 90◦m + 270◦n = 180◦(k − 2) =

180◦(m + n− 2). Solving this, we obtain m = n + 4.

It follows that any finite rectilinear polygon must have at least 4

convex corners. If the region is not finite, it may not have any cor-ners. And if we allow holes, we can have more concave corners thanconvex corners.

18 polyominoes

Problem 4.

(1) Give examples of regions (not necessarily finite or simply-connected)with:

(a) No corners.

(b) One convex corner and no concave corners.

(c) Two convex corners and no concave corners.

(d) Four concave corners and no convex corners.

(2) Prove the following regions are impossible. Regions with:

(a) Three convex corners and no concave corners.

(b) One, two, or three concave corners and no convex corners.

(3) Prove that all other combinations are possible.

Problem 5. Prove that (up to symmetry) there is a unique rectilinear figurewith 6 sides whose side lengths are given.

Example 3. Suppose we can place n 2× 2-squares anywhere on the lattice.They are allowed to overlap but not coincide. Then they must cover at leastn + 3 cells.

Here is why: Any cell can be covered by at most 4 squares. Convexcorners can only be covered by one square. There must be at least 4 con-vex corners (Theorem 8). So all possible cells that can potentially be cov-ered by 4 squares is given by 4n−4

4 = n − 1. So there must be at leastn− 1 + 4 = n + 3 cells in total.

Figure 15: In this region, convex cornersare red, and concave corners are blue.The five mountains that lie betweenpairs of convex corners are markedred, and the valley between the pairof concave corners is blue. The tworemaining (black) edges are flats.

An edge is called a mountain if lies between two convex corners, avalley if lies between two concave corners, and a flat otherwise.8

8 In Bar-Yehuda and Ben-Hanoch (1996)the author calls a mountain a knob anda valley an anti-knob. The term peakis used in Beauquier et al. (1995) fora similar concept, so I chose to usethe term mountain, since it allows thenatural extension to valleys and flats.

Theorem 9 (ccorn (2017) via Wikipedia contributors (2017)). Let Mbe the number of mountains and V the number of valleys of a finite simply-connected region. Then M−V = 4.


Proof. Let nx be the number of convex corners, and ny the number ofconcave corners. Let nxx be the number of convex corners followedby a convex corner, nxy the number of convex corners followed by aconcave corner, and so on.

Then nx = nxx + nxy = nxx + nyx and ny = nyx + nyy = nxy + nyy,which means nxx = nx − nxy = nx − (ny − nyy) = nyy + (nx − ny). Butby Theorem 8, nx − ny = 4, so nxx = nyy + 4.

It follows that for finite simply-connected regions, the number ofmountains must always be at least 4. Furthermore:

introduction 19

Theorem 10. If F is the number of flats for a finite simply-connected re-gion, then F is even.

[Not referenced]

Proof. Since M − V = 4 (Theorem 9), it follows that M and V musteither both be odd or both be even. So their sum is even. But M +

V + F = k, and k is even (Theorem 7), therefor F must be even.

The following theorems, although they seem intuitively true, re-quire a more precise definition of both polyomino and hole. I statethem here for completeness; see the reference for details.

Theorem 11 (Herzog et al. (2014), Lemma 1.1). Every hole is simply-connected.

[Not referenced]

Theorem 12 (Herzog et al. (2014), Lemma 1.2). In a simply-connectedpolyomino, if a vertex is shared between exactly two cells, the cells are neigh-bors; that is, they cannot be diagonally opposite.


Theorem 13. For a rectilinear polygon, the number of holes H is related tothe number of mountains P and valleys V by the following equation:

H =V −M

4+ 1.


Proof. Suppose we have holes 1, 2, 3, . . . , H, and the border of hole ihas Mi mountains and Vi valleys, and the outer border of our poly-gon has M0 mountains and V0 valleys. The total mountains andvalleys is given by:

M =H

∑i=0

Mi and

V =H

∑i=0

Vi.

And so

V −M =H

∑i=0

(Vi −Mi).

If we consider the border of a hole as a rectilinear region on itsown, we notice that its mountains correspond to valleys in the origi-nal polygon, and its valleys corresponds to mountains in the original

20 polyominoes

polygon. So it has Vi mountains and Pi valleys, and so we know byTheorem 9 that Vi −Mi = 4, for i > 0. For i = 0 we have M0−V0 = 4,and so we get the following:

V −M = V0 −M0 +H

∑i=1

(Vi −Mi)

= −4 +H

∑i=1

4

= −4 + 4H

= 4(H − 1),

and soH =

V −M4

+ 1.

A lattice polygon is a polygon whose vertices fall on a lattice.

Theorem 14 (Pick’s Theorem, Niven and Zuckerman (1967), Theorem4). The area A of a lattice polygon, is given by A = b/2 + r− 1, where b isthe number of lattice points on the boundary of the polygon, and r is numberof interior vertices.9 9 This theorem was discovered by Pick

(Pick (1899)); Bruckheimer and Arcavi(1995) provides some history andcontext; more references are given inthe further reading section.


Proof. The details of the following outline is easy to fill in:

(1) The theorem is true for rectangles.

(2) The theorem is true for right triangles.

(3) The theorem is true for triangles.

(4) The theorem is true for polygons.

For details, see the reference.

Theorem 15 (Williams and Thompson (2008), Theorem 1). If thenumber of interior vertices of a simply-connected n-omino is given by r, theperimeter p is given by:

p = 2n− 2r + 2.


Proof. We use induction on n.The theorem is true for the monomino and domino.Suppose then it is true for a polyomino with n cells and r interior

vertices, so that the perimeter p = 2n− 2r + 2.

introduction 21

We can attach another cell to the polyomino to form a new poly-omino with n + 1 cells. Let p′ be the perimeter of this polyomino, r′

be the number of interior vertices of the new polyomino. We want toshow that p′ = 2(n + 1)− 2r + 2.

There are four possibilities (shown in Figure 16). (a) Case 1. (b) Case 2.

(c) Case 3. (d) Case 4.Figure 16: The cases of Theorem 15.The blue cells are from the old poly-omino, and the yellow cell is added toform the new polyomino.

• Case 1: No new interior vertices are added. The new perimeteris 2 units longer; so p′ = p′ + 2 = 2n− 2r + 4 = 2(n + 1)−2r′ + 2.

• Case 2: The perimeter stays the same, but one more interiorvertex is added. So p′ = p = 2n− 2r + 2 = 2(n + 1)− 2(r +1) + 2 = 2(n + 1)− 2r′ + 2.

• Case 3: In this case two new interior vertices are added, andthe perimeter is reduced by 2. Thus p′ = p− 2 = 2n− 2r + 2 =

2(n + 1) + 2(r− 2) = 2(n + 1)− 2r′.

• Case 4: This case is actually impossible if the original poly-omino is simply-connected by Theorem 12.

The proof is also easily derived from Pick’s Theorem (Theorem14), by noting that for polyominoes, n = A and b = p (Williamsand Thompson, 2008). This theorem gives us another proof that theperimeter is even as was proved in Theorem 5. The following is anextension of the theorem above to polyominoes with holes.

Theorem 16. For a polyomino with H holes, the perimeter is given by:

p = 2n− 2r− 2H + 2.

[Not referenced]

Proof. Let P′ be the polyomino with all the holes filled.Let p0 denote the outer perimeter, and pi the perimeter around

hole i. Let n0 denote the number of cells in the polyomino plus all thecells in the holes, and ni the cells in hole i. Let r0 be the number ofinterior vertices of P′, and ri the interior vertices of hole i.

The number of vertices on the perimeter of a polyomino is thesame as the length of the perimeter of the polyomino. Therefor r =

r0−∑Hi=1(ri + pi). We also have n = n0−∑H

i=1 ni and p = p0 +∑Hi=1 pi,

and pi = 2ni − 2r2 + 2 by Theorem 15.Let us expand p0 −∑H

i=1 pi:

22 polyominoes

2n− 2r− 2H + 2 = 2(n0 −∑ ni)− 2(r0 −∑(ri + pi))−∑ 2 + 2(2.1)

= (2n0 − 2r0 + 2) + ∑ [−2ni + 2ri − 2 + 2pi] (2.2)

= p0 + ∑ [−pi + 2pi] (2.3)

= p0 + ∑ pi (2.4)

= p. (2.5)

Problem 6.

(1) Do any of the polyominoes with n cells and minimum perimeter haveholes?

(2) Can you characterize the minimum perimeter polyominoes?

See also Problem 10.

Theorem 17 (The perimeter criterion). Suppose we have a region withmountains of length e1, e2, . . . and a set of tiles with mountains of lengthe′1, e′2, . . .. A necessary condition for the region to be tileable is that eachmountain of length ei, we have ei = ∑j nje′j for some ni ≥ 0.

[Referenced on pages 154, 155, 156 and 158]

Example 4. The 9 × 9 square with the center removed cannot be tiledby 2 × 2 squares. The area of the region is 80, and so it satisfies the areacriterion. However, we cannot express the edge of length 9 as a multipleof 2, and so it fails the perimeter criterion. In fact, we cannot tile a 9× 9square with a cell removed from anywhere, since we always have at least twoborders of odd length. Surprisingly, the minimum number of monominoesnecessary to to tile a 9× 9 square is actually 17! (See Theorem 173.)

Theorem 18 (The row-criterion). Suppose our region rows of length ei

and our tiles have rows of length e′i . If no rotation is allowed, then a tilingcan exist only if ei = ∑j nje′j for some ni ≥ 0.


Example 5. The tile in Figure 5 can only tile regions with all rows andcolumns containing an even number of cells. (In fact, if a row or column isnot connected, each piece must have an even number of cells.)

Figure 17: A tile that can only tileregions that have an even number ofrows and columns.

The following fact is used to show that if we rotate a polyomino90°around a point and the rotated copy does not intersect the orig-inal, then a copy rotated 180° also does not intersect the original.

introduction 23

(Later, we use this to put constraints on how certain polyominoes canfit into rectangles.)

This fact seems reasonable, but requires some ideas from topologyto prove, and therefor I omit the proof.

Theorem 19. Suppose P and Q are points such that P is Q rotated about180° around O, and there is a continuous path p between P and Q. Supposewe rotate p 90° around O to form a new path p′. Then p and p′ intersect.


Figure 18: A curve between two pointsrotated intersects itself under certainconditions.

For a proof, see von Eitzen (2019). The Further Reading section hassome references to related ideas.

Theorem 20. Suppose that P0 is a polyomino, and we rotate it about anypoint 90°, 180° and 270°degrees to get P1, P2 and P3. If P0 ∩ P1 is empty,then Pi ∩ Pj is empty for all i 6= j.


Proof. If P0 does not overlap P1, then by symmetry P1 does not over-lap P2; P2 does not overlap P3; and P3 does not overlap P0. We willshow P0 does not overalp P2; the remaining cases follow by symme-try.

Let’s assign coordinates such that the center of rotation is at (0, 0),one cell is a unit square, with its edges aligned to the axes.

Note, the center of rotation can fall on either a cell center, or avertex; it does not affect the proof.

Also note that if the center of rotation falls on a cell, that cell can-not be part of P0, since otherwise it will also be part of P1, whichcontradicts the hypothesis.

Now suppose that P0 overlaps P2. This means there is are twocells (x, y) and (−x,−y) that are part of P0. Since P0 is connected,there is a path from (x, y) to (−x,−y). By Theorem 19 this pathmust intersect with a rotated copy, which means P0 must intersect P1,which is impossible. Therefore, P0 cannot overlap P2.

2.4 Further Reading

Golomb (1996) and Martin (1991) are the classical references on poly-ominoes; both focus on their recreational aspects and do not requiremuch mathematical background. Toth et al. (2017, Chapter 15) givesa broad overview of polyominoes, and Soifer (2010, Chapter 1) dis-cusses several problem-solving techniques and contains various poly-omino tiling exercises10. Guttmann (2009) is an advanced book, and 10 Chapter 5 is a note that also deals

with a polyomino tiling problem.deals with the types of problems that have applications in physics.

24 polyominoes

Winslow (2018) gives a list of interesting open problems concern-ing polyominoes.

There is a substantial number of web sites dealing with poly-ominoes. Below are the ones I think are most useful for referencepurposes; they all contain lots of data.

• Grekov lists the number of known tilings of rectangles byvarious polyomino tile sets.

• Reid contains lists of prime rectangles for polyominoes thattile rectangles.

• Dahlke contains some tilings of rectangles by various poly-ominoes, as well as analysis for polyominoes that don’t tilerectangles.

• Myers contains lists of the number of polyominoes that can tilethe plane in various ways.

• Oliveira e Silva (2015) contains lists of counts of polyominoes,broken down by the area of their holes and symmetries.

The magnificent Grünbaum and Shephard (1987) covers the gen-eral topic of tiling in depth. For a lighter overview, see Ardila andStanley (2010). The lecture notes Bassino et al. (2015) covers manytopics using ideas similar to the ones in these essays. There are morereferences for general tiling in Section 7.7.

For more on rectilinear polygons, see O’rourke (1987) and Preparataand Shamos (2012). For more on Pick’s Theorem, see Varberg (1985)and Grünbaum and Shephard (1993). For a more extensive list of ref-erences on Pick’s Theorem, see Dubeau and Labbé (2007, Section 1).A very nice application of Pick’s Theorem to solve a polyomino com-patibility problem is given in Sayranov. (Polyominoes are compatibleif they can tile the same region).

I asked for a proof of Theorem 19 on Mathematics StackExchange(see Tulleken (2019)). This question prompted and uncovered severalrelated questions. See YiFan (2019), Gilhooley (2019a), Gilhooley(2019b) and Tremblay (2019).

If you read Latvian, then Cibulis (2001a) and Cibulis (2001b) con-tains a large number of pentominoes puzzles with solutions andrelated information.

introduction 25

2.5 Hexominoes, and Heptominoes

This section shows the hexominoes and heptominoes, and is includedfor reference purposes.

Figure 19: Hexominoes

26 polyominoes

Figure 20: Heptominoes

Figure 21: Heptominoes (Continued)

Figure 22: Heptominoes (Continued)

3Dominoes I

Figure 23: A tiling of a 8× 8 square bydominoes.

Figure 27 shows some examples of regions that cannot be tiledby dominoes. That some regions cannot be tiled by dominoes isintriguing. How can shapes as small as dominoes not be arrangedto fit into a region as big as in Figure 27(e), when there are so manyoptions?

This is the topic of the first section: understanding why someregions have a tiling by dominoes and others don’t.

Figure 24: The mutilated chessboard.

Figure 68 provides us with another mystery. In each region, thereare dominoes that fit into the tiling in just one way (they are markedin yellow). In the first region it is easy to see why this must be so,but what is happening in the last region? How is it not possible tofind a tiling so that at least some of those dominoes lie in a differentposition?

This is the topic of the second section: understanding how thesame region can be tiled in different ways and how the region canforce dominoes into certain positions.

A famous problem, the mutilated chessboard problem1, illustrates 1 The mutilated chessboard problem,was first proposed by Max Black inBlack (1947), and has been discussedin various places, including (Golomb,1996, p. 4), (Martin, 1991, p. 1-4, 7-9),(Mendelsohn, 2004) and Engel (1998).

some key ideas from each section.Consider an 8× 8 chessboard, with two opposite corners removed

as in Figure 24. Can the board be tiled? If you know this problem,you know that it cannot. Because, every domino must cover exactlyone black and one white square, and with opposite corners removed,there are more cells of one color than the other, and so a tiling isimpossible. Of course, this principle can be applied to any region,and gives us a valuable tiling criterion. We will use this as the basisfor constructing tiling criteria more powerful than the area criterion(Theorem 1) that we gave in the introduction.

Figure 25: Dividing the chessboard intotwo strips

If we remove two squares of opposite colors instead, is a tilingalways possible? The answer is, yes, a tiling is always possible. Herethey key is to notice that we can divide the board into two strips, as

28 polyominoes

shown in Figure 25. And because the cells have different colors, nomatter where we remove the two cells from, the end-points of eachstrip must have different colors, and so it has an even number ofcells. Each of these strips has an even number of cells, and is tileablein the obvious way. (We will go over this logic in more detail once wemade some proper definitions.)

One way to show a region is tileable, is to partition it into stripswith an even number of cells. If the end point of a strip of morethan two cells is next to its starting point, we say the strip is closed.Closed strips have at least two tilings, and this is the basis of thesecond section. We will see that cells that can only be tiled one waycan never be part of a closed strip, and that all tilings of a region canbe obtained from the two tilings of each closed strip in it.

Figure 26: A tiling along the strips.

3.1 Tiling Criteria

In this section we develop some tools with which we can tellwhether a region is tilable by dominoes or not.

We could, of course, use brute force to check all the regions inFigure 27 (or get a computer to do it). There are two reasons to lookfor something better:

• So that we can write faster programs. While a naive programmight deal with all our examples in seconds, scientists usingtilings to understand how molecules fit in solids need some-thing better. Their “tiling problems” may be regions withmillions of cells!

• So that we can understand how these tilings work, and un-derstand other related phenomena. Section 3.2 gives a taste ofthis.

Our goal is to develop four techniques:

• A flow criterion: a technique of coloring cells in a region andanalyzing the counts of dominoes that must cross the bordersof subregions.

• Cylinder reduction: a geometric transformation that preservesthe tileability of a region and can be used to simplify regions.

• The marriage theorem: another way to determine if a tilingexists.

• A generalized way of coloring regions to exhibit their untili-ablity.

dominoes i 29

(a) An exampleof an unbalancedpolyomino.

(b) An exampleof a balancednon-compact poly-omino that cannotbe tiled.

(c) An example of a balancedcompact shape that cannot be tiled.

(d)

(e)

(f)(g) (h)

(i) No matter how thisregion and the next one areextended at the blobs, theresulting regions will notbe tileable

(j)

Figure 27: Regions that cannot be tiledwith dominoes.

30 polyominoes

We do not arrive at the state-of-the-art in solving domino tilingproblems (just yet), however, the tools will help us deal with all theregions in our examples in Figure 27, and give us an intuitive under-standing of how to tackle tiling problems.

A key point to understand from this section is: the border of aregion is important, and tell us a lot about how a region can be tiled.

3.1.1 Flow

Since all tilings must satisfy the area criterion (Theorem 1), weknow a tiling by dominoes can exist only if the area of the region wewish to tile is even. Not all regions with even area are tileable (all theregions in Figure 27 have even area).

Now suppose we divide a region into two parts, each with an oddarea. If the original region is tileable, then we know that there mustbe at least one domino that lies in both of the two parts. In fact, weknow the number of dominoes that crosses the border between thetwo parts must be odd, otherwise the remaining regions will haveodd area and not be tileable. Figure 28 illustrates this principle.

Figure 28: Two examples of the tilingof a square. The square is divided intotwo partitions (yellow and blue). Sinceboth partitions have odd area, any tilingsuch as the two shown above must havean odd number of dominoes that crossthe border.

Theorem 21 (Border Crossings Theorem). In a subregion S of a regionwith a tiling, the number of dominoes that cross the border of the subregionmust have the same parity2 as the area of the subregion.

2 Even or odd.[Referenced on pages 31, 34, 53, 91, 112 and 119]

Proof. Let k be the number of dominoes that cross the border of S.Each of these dominoes has only one cell in S. If we remove thesecells to form a new region S′, we have a region that is tileable, witharea |S′| = |S| − k. Since S′ is tileable, |S′| must be even (by Theorem1), and so |S| and k must either both be odd, or both be even.

Below we give two applications of this theorem; as a theorem andan example.

A bridge is a sequence of cells in a region such that each has ex-actly two neighbors, and such that removing any cell in the bridgewill leave the region disconnected. For example, Figure 27(f) and(g) each has a bridge, Figure 27(h) does not (since no cell splits theregion when removed). Figure 27(c) also does not have a bridge; al-though there are single cells that will split the region if removed, theyhave more than two neighbors.

Theorem 22. Each cell in a bridge can be tiled in only one way. 3 3 Such cells are called frozen. We willdefine this concept in the next section.

[Not referenced]

dominoes i 31

Proof. Let R be a region with a bridge, and let v be a cell in thatbridge. Because v is a cell in a bridge, it has exactly two neighbors–let’s call them u and w. Further, R − {v} is disconnected. Let Su

and Sw be the two disconnected subregions that contains u and wrespectively. The border of each of these share exactly one edge withthe cell v, in particular, Su shares a border with v at the edge betweenu and v.

Now consider Su. If |Su| is odd, then the number of dominoes thatcrosses the border of Su is odd. The only place where a domino cancross is for a domino to cover both u and v, and therefor any tiling ofR must have a domino in this position.

If on the other hand S|Su| is even, then the number of dominoesthat crosses the border of Su is even. Since there is only one placewhere a domino can cross, it means the number of dominoes thatcross must be 0. Therefor, a single domino cannot cover both u and v,and so, a single domino must cover v and w. Therefor, every tiling ofR must have a domino in this position.

Taken these together, v can only be tiled one way. The same argu-ment applies to all oither cells in the bridge, and therefor the bridgehas a unique tiling in R.

Figure 29: The cells marked yellow canbe tiled in only one way.

The next example uses Theorem 21 in a more sophisticated way toprove that a tiling does not exist. 4

4 The author mentions this idea hasbeen known before.Example 6 ((Mendelsohn, 2004)). Let’s look at the mutilated chessboard.

The top row has an odd area. Therefore, an odd number of dominoes mustcross its border (Theorem 21), and only vertical dominoes that also lie in thesecond row can do that (otherwise, parts of dominoes will fall outside themutilated chessboard).

The second row has even area, and so an even number of dominoes mustcross its border. We already have an odd number of dominoes that crossthe border from the top, therefore we must have an odd number of verticaldominoes that cross the border to the bottom.

Following the same argument, we get that there must lie an odd numberof vertical dominoes between each pair of adjacent rows. There are 7 suchpairs, so we can conclude the total number of vertical dominoes must be odd.

Applying this idea to the columns, we find there must be an odd num-ber of horizontal dominoes too. So the total number of dominoes must beeven. But to cover the 62 squares we need 31 dominoes, which is odd. Andtherefore, a tiling is impossible.

This example shows that we can always determine the parity ofthe number of vertical and horizontal dominoes in a tiling. For atiling to exist, this must be consistent with the parity of the totalnumber of dominoes (which we can deduce from the area).

32 polyominoes

Satisfying the area criterion is not enough (all the regions inFigure 27(a)-(h) are untileable and have even area), and checking theparity as in the previous example is also not enough (for example,Figure 27(f)).

We will now look at the color argument we discussed at the be-ginning of the chapter, and see how far it gets us. The checkerboardcoloring plays an important role in our discussions going forward,and to make it easier to talk about and prove details we introducesome additional notation and terminology.

In a region R with checkerboard coloring applied, letW(R) denotethe white cells in R, and let B(R) denote the black cells in R. Thedeficiency5 of R is defined as 5 Also called bias.

∆(R) = |B(R)| − |W(R)|. (3.1)

If the deficiency of a region is 0, the region is balanced.6 6 The term balanced is used informallyin (Golomb, 1966, p. 17). The term , aswell as the white and black functions,are defined explicitly in for exampleThiant (2003). The term is also used inHerzog et al. (2015) for a completelydifferent concept. The words biased andunbiased is sometimes used instead ofunbalanced and balanced, for exampleMason (2014).

The functionsW , B and ∆ depend on which of two ways the im-age has been colored. However, the absolute value of the deficiencyand being balanced are inherent features of a region, and are inde-pendent of the coloring used.

Theorem 23 (Checkerboard Criterion). For a region to be tileable bydominoes, it must be balanced (Golomb, 1996, p. 4).


Problem 7.

(1) Prove that ∆(R) ≡ |R| (mod 2). Note that this imploes that thearea criterion is redundant.

(2) Show that ∆(R(m, n)) equals 0 or 1 depending on whether the areaof the rectangle is odd or even.

(3) Prove |∆(R)| ≤ |R|.

(4) Prove that if R is partitioned into two subregions S1 and S2, then|∆(R)| ≤ |∆(S1)|+ |∆(S2)|.

Using this criterion, we can prove Figure 27(a) is not tileable.However, it still is not enough, since all the regions in Figure 27(b)-(h)are balanced and untileable.

Problem 8. Find examples of non-tileable balanced regions.

What is the largest the deficiency can be? Figure 30 shows we canmake the deficiency as large as we want by extending it as shown.However, it is bounded by the number of cells, as stated in the fol-lowing theorem.

dominoes i 33

Theorem 24. The deficiency of a connected region R is bounded by thenumber of cells as follows:

|∆(R)| ≤ |R|+ 12

.

[Not referenced]

Proof. Let B = B(R) and W = W(R). WLG, assume that B > W,so ∆(R) > 0. We will shows that B ≤ 3W + 1, from which the resultfollows. Because if B ≤ 3W + 1, we have 2B− 2W ≤ W + B + 1, thatis, 2∆(R) ≤ |R|+ 1, or ∆(R) ≤ |R|+1

2 .We now prove B ≤ 3W + 1. Suppose not; that is, suppose B >

3W + 1. Then there must be at least one white cell with 4 blackneighbors, with none of these 4 black cells having any other whiteneighbors. If there are no other cells than these 5 cells, we have4 = B = 3W + 1, so there must be more. But then these 5 cellscannot be connected to any other cells, and we have a disconnectedregion which contradicts our hypothesis that the region is connected.

Figure 30: A family of regions thatobtain maximum deficiency.

This upper bound is achievable by regions like the one shown inFigure 30. A polyomino that achieves the maximum amount of biaspossible for its area is called maximally biased (Mason, 2014). Thenumber of maximally biased polyominoes is shown in Table 4.

n A234013

1 1

2 1

3 2

4 1

5 1

6 11

7 8

8 3

9 1

Table 4: The number of maximallybiased polyominoes.

|R| |∆(R)|

4k 2k4k + 1 2k + 14k + 2 2k4k + 3 2k + 1

Table 5: Maximum deficiency.

Problem 9. (Mason, 2014) Show the following facts about 4n + 1-sizedbiased polyominoes, with B(R) >W(R):

• They are all of made from a monomino and T-tetrominoes like the oneshown in Figure 30.

• The removal of any white square will result in an illegal, discon-nected polyomino.

• The same for any twice-connected black square.

• The removal of any once-connected black square will result in a legal,maximally biased polyomino of size 4n.

• The addition of a white square will result in a maximally biased(4n + 2)−polyomino;

Not all maximally biased (4n + 2)-polyominoes can be generatedfrom (4n + 1)-maximally biased polyominoes, as shown in Figure 31

(Mason, 2014).

Problem 10.


34 polyominoes

(1) Show that the maximum perimeter for a polyomino with area n is2n + 2. (See also Theorem 15).

(2) Show that the number of polyominoes with area n that have aperimeter of 2n + 2 (A131482) is the number of maximally biasedpolyominoes for n = 4k + 1.

See also Problem 6.

Figure 31: A maximally biased poly-omino that cannot be formed by ex-tending a maximally biased polyominowith one less cell.

In the same way that we turned the area criterion into somethingmore useful by looking at what must happen at the border of subre-gions, we now turn the color criterion into something more powerfulby looking at the border of subregions.

Let R be a region with the checkerboard coloring applied, andlet S be a subregion of R. If w is the number of dominoes coveringthe border of S with a white cell inside S, and b is the number ofdominoes covering the border with their black cells inside R, then wedefine the flow7 of S as 7 This definition is essentially given in

Saldanha et al. (1995).

φ(S) = b− w. (3.2)

See Figure 32.When S = R, the flow is 0, since there are no dominoes that cross

the border, and so w = b = 0.

Figure 32: Let S be the region inside thered border. Then b = 2, w = 1, andso φ(S) = 2− 1 = 1. Also, B(S) = 5,W(S) = 4, and so ∆(S) = 5− 4 = 1.

Theorem 25 (The Flow Theorem). Let S be a subregion of R, and supposeR has a tiling by dominoes. Then the flow of S equals the deficiency of S,that is,

φ(S) = ∆(S). (3.3)


Proof. Let W (B) be the number of white (black) cells inside S, letw (b) be the number of dominoes that cross the border with theirwhite (black) cells inside S, and let k be the number of dominoescompletely in S. Then B−W = (b + k)− (w + k), and so B−W =

b− w, and thus ∆(S) = B−W = b− w = φ(S).

Note the similarities between border crossings theorem (Theorem21) and the flow theorem above. Both give us information about whathappens at the border based on what is inside the region. But theflow theorem gives us much more information; Theorem 21 merelytells us the parity of the number of dominoes that cross the border.The flow theorem gives us the difference of the two different types ofdominoes.


dominoes i 35

Also note that the area must have the same parity as the flow: ifthe area is even, then W and B are both odd or both even. In eithercase, their difference is even. This means with the flow theorem inplace, the border crossing theorem is now redundant. 8 8 One reason to differentiate the two

theorems is that the border crossingstheorem is easier to generalize to othertile sets.

The following examples show how to use the flow theorem toprove a region is not tileable.

Example 7. Consider Figure 27(c), and let’s choose the subregion as theshape left of the dotted line. Now since W = 3 and B = 4, we have|W − B| = 1, so we know a domino must cross the dotted line. Removingthis domino partitions the shape into two shapes, each of which is untileablebecause they don’t satisfy the area criterion.

Example 8. Consider Figure 27(d), and choose the subregion as the regionthe right of the dotted line. We have |B−W| = 2, which implies dominoesmust overlap the dotted line in two places. But this means b = w = 1, andso |b− w| = 0, which is impossible if the region is tileable. Therefore, it isnot tileable.

Example 9. Consider Figure 27(e). Partition it in halves by a verticalcut through the middle. Then |B −W| = 4, but the maximum value that|b− w| can have is 3. Therefore, the region is not tileable.

Example 10. Consider Figure 27(i), and consider the colored subregion.The deficiency |B−W| = |8− 5| = 3. This means, at least three dominoesmust overlap the border. However this is done, we are always left with aregion with 4 black and 3 white squares, which is untileable. Therefore, theentire region is untileable.

Problem 11. If we have a domino tiling iof a region, show that we can de-termine the parity of vertical dominoes with their top squares black. (Followthe type of reasoning used in Example 6.)

If you tried your hand at Problem 8 and followed the examplesabove, you may have noticed that you can create an untileable regionby having an abundance of black on the one side of the region, and aabundance of white on the other side, with a choke point between thetwo parts. The following theorem gives a formulation of this idea.

Figure 33: The deficiency of the leftpartition is -2, so the length of a cutthrough the center must be at least 3.

Theorem 26. 9 Suppose we apply the checkerboard coloring to a tileable

9 In Kenyon (2000c) the author mentionsthat a very similar theorem has beenproven in Fournier (1996) (in French).

region, and partition it into two subregions with a straight cut. If one sub-region has W white cells and B black cells, then the cut must have length atleast 2|B−W| − 1.


Proof. Assume B ≥ W. From the flow theorem (Theorem 25), weknow that the number of tiles that cross the cut must be at least B−

36 polyominoes

W = b− w. From this we get b = B−W + w, so b ≥ B−W (since wis non-negative). So the number of dominoes that cross the cut witha white cell inside the subregion must be at least B −W. Along a

straight cut, there are at most⌈

L2

⌉places where this can happen, so it

needs to have length at least 2(B−W)− 1.If we assume W ≥ B, we can show the length of the cut must be at

least 2(W − B)− 1 following the same argument as above, reversingthe roles of black and white.

Putting these together, we arrive at the result: the length of the cutis at least 2|W − B| − 1.

See Figure 33 for an example.

Problem 12. What if the cut is not straight?

You may also have noticed that to create unbalanced regions orparts of regions, you have to manipulate the border of the region soyou have a lot of corners of the same color. Also, we have alreadyseen some theorems that relate the border of a region to what isgoing on inside. The following theorem shows we can determine thedeficiency from what is going on at the border alone.

Theorem 27. Let b be the number of black edges on the border, and w be thenumber of white edges on the border. Let B be the number of black squaresand W be the number of white squares. Then b− w = 4(B−W) = 4∆(R).


Proof. Consider building a region cell by cell. At each stage, we caneither add a white or a black square. If we add a white square, it isa neighbor of 0, 1, 2, 3, or 4 other cells, all of which must be black.Each exposed edge must be white, and each unexposed edge mustreduce the total number of black edges. So the total amount thatb − w decreases is 4. A similar argument shows that if we add ablack square, b − w is increased by four. In other words: b − w =

4(B−W).

It follows that B = W if and only if b = w.

Problem 13. A corner cell is a cell with two adjacent edges that are notshared with other cells. Let R be a region such that each of its corner cellshave no neighbors that are on the border (that is, all neighbors of each cornercell are interior cells). Prove R is untilable.

Theorem 28 (Csizmadia et al. (1999), Theorem 2.1, Theorem 3.1). If alln edges of a simply-connected region R have odd length,

dominoes i 37

(1) the deficiency of R is given by

∆(R) =n4

,

(2) the region is unbalanced, and

(3) the region is untilable by dominoes.


Proof. 10 Suppose one corner edge is black. Then all corner edges are 10 The proof here is new. The proofsin Csizmadia et al. (1999) is quitecomplicated.

black. This means the ends of sides are all black, and so for each sidei, we have:

Bi −Wi = 1,

and summing over all sides:

n

∑i=1

(Bi −Wi) = n.

But ∆(R) = ∑ni=1(Bi−Wi)

4 (by Theorem 27), and thus ∆(R) = n4 > 0,

and so R is unbalanced, and hence untilable by dominoes. Note wealready proved that n is divisble by 4 in Theorem 6.

The theorem does not hold for regions with holes. For example,the 3× 3 square with its center removed has all its sides odd, but it isbalanced and even tileable (Figure 34).

Figure 34: A region with a hole thathas all its edges with odd length that isbalanced and tileable by dominoes.

Theorem 29. A balanced polyomino must have at least two sides of evenlength.


Proof. For the polyomino to be balanced, it must at least have oneeven side (Theorem 28). We also know the perimeter must be even(Theorem 5). Therefor, the number of odd sides must be even. Butthe total number of sides must be even (Theorem 7), and so the num-ber of even sides must also be even. Thus, there must be two or moreeven sides.

Theorem 30. A polyomino with all sides even is tileable. (Mentioned inKenyon, 2000b, Section 8).


Proof.

(1) For regions with no holes:

38 polyominoes

(a) If the figure has exactly four knobs, it must be a rectangle.In this case we can tile the region with horizontal dominoesonly.

(b) If the figure has more than four knobs, we can remove one.Suppose a knob B lies between sides A and C, with A ≤ C.We can remove a rectangle S1 with sides of length B and A,and since both are even we can tile S1 with horizontal domi-noes. We can repeated this process until a single rectangle isleft, which we can tile as we did above.

Note that the long edge of a domino is shared with at mostone domino.

(2) For regions with holes:

(a) First tile the filled polyomino (i.e. the polyomino with allholes filled) with horizontal dominoes as in the procedureabove.

(b) Remove all dominoes that lie completely inside a hole.

(c) The ones that lie partially in a hole come in pairs, sinceall sides are even. Each pair can be replaced with a singledomino that covers only the cells that do not lie in the hole.

This completes the tiling.

Figure 35: An example showing theprocedure in the proof.

Problem 14.

(1) Can you modify the theorem’s conditions to be weaker?

(2) Prove that if the sides of a region is divisible by n, it can be tiled bybars of length n if holes are at least n− 1 units apart.

(3) If we use all n-ominoes for the tileset, can holes be anywhere?

dominoes i 39

3.1.2 The Marriage Theorem

Informally, the marriage theorem states the following11: Suppose 11 This is a specific version of a muchmore general theorem given in Hall(1935) that is now called the marriagetheorem, with applications that gobeyond tiling. Covering the moregeneral results fall outside the scope ofthis essay, but we give some referencesin the Further Reading section. Thisversion is given in Ardila and Stanley(2010).

we have a group of k white cells from a region R, and they have nblack neighbors in R. If n < k, then no tiling exists. Moreover, ifevery group of white cells in R has at least as many neighbors aswhite cells in the group, a tiling exists.

Before we prove it, we need some terminology to make it easier tomake our statements.

A subregion S of R is called a white patch of R if all the neighborsof its white cells are also in S, and there are no other black cells in S.In other words, all black cells in S have at least one neighbor also inS. A similar definition can be made for black patch.

Figure 36: In this region, the cellssurrounded by a pink line is a blackbad patch. The rest of the region is awhite bad patch.

A region itself is a patch. If a patch is not equal to the region wecall it a proper patch. A white patch can be, but is not necessarily,a black patch. In fact, if R is connected, a white patch is only also ablack patch if it is the entire region.

Theorem 31. Suppose R is connected, and S is both a white patch and ablack patch. Then S = R.

[Not referenced]

Proof. Suppose R has some cells that are not in S. Since R is con-nected, we must have at least one of these cells be a neighbor to a cellin S. Let’s call this cell u, and its neighbor in S, v. Suppose v is black,then because S is a black patch, all its neighbors must lie in S, andthis contradicts that u lies outside S. And if v was white, because it isa white patch, all its neighbors must lie in S. Therefore, there can beno cells in R that are not also in S, and therefore S = R.

Theorem 32. Let S be a white patch of a region R. Then R− S is a blackpatch of R. Similarly, if S is a black patch of R, then R− S is a white patch.


Proof. We only prove the first part; the second part can be provenwith the exact same argument with the roles of white and blackreversed.

Suppose S is a white patch, but R − S is not a black patch. Thenthere is a black cell u in R− S with a neighbor v not in R− S. Then vmust lie in S, and it is white. Since S is a white patch, all the neigh-bors of v, including u must lie in S. We arrive at a contradiction, andso R− S must be a black patch.

40 polyominoes

We call a white patch bad if it has fewer black than white neigh-bors, and good otherwise. Similarly for black patches.

Theorem 33. A balanced region has a bad black patch if and only if it has awhite bad patch.


We only prove the first part, the second part follows from the sameargument reversing the roles of white and black.

Proof. If. Suppose that S is a bad white patch. Then R− S is a blackpatch (Theorem 32). Since S has fewer black than white cells and R isbalanced, R− S must have fewer white cells than black cells, and so itis bad.

It follows directly from this that if all white patches in a region aregood, then so are all black patches.

Theorem 34 (The Marriage Theorem). A region is tileable if and only ifit has no bad patches.


Proof. If.12 Suppose all the white patches of a region are good. It 12 The logic of this part of the proof fol-lows the proof in (Kung et al., 2009, p.56), due to Easterfield (Easterfield, 1946)and Halmos and Vaughan (Halmos andVaughan, 2009).

follows that all the black patches of the region must be good too (andvice versa) by Theorem 33. We can now partition the region into twosubregions S and R− S in one of two ways:

(1) If all the proper white patches of R are unbalanced, then eachpatch of R must have strictly more black cells than white cells.Let S be a white cell and its neighbor. Then S is tileable (witha single domino), and R − S is a region with all its whitepatches good. This follows from the fact that we removedonly one black cell, so the number of black cells in each patchcan can drop by at most 1, and since these are strictly biggerthan the number of white cells, after the drop there must be atleast as many black cells as white cells in the patch.

(2) If at least one of the patches in R is balanced, we let S be sucha patch. Then S must have all its white patches good, andR− S must have all its black patches good (and so, it must alsohave all its white patches good). Both partitions must also bebalanced.

Note that a patch in S or R− S need not be a patch in R.

dominoes i 41

We can continue this process, and it must eventually end, since thenumber of cells in R is finite. And so, we eventually arrive at a bunchof subsets of two cells each, all tileable by a single domino, and so theentire region must be tileable by dominoes.

Only if. If the region is not balanced, we know it is not tileable byTheorem 23.

Suppose then it is balanced, and suppose it has a bad white patch.(If it had a bad black patch, it must also have a bad white patch byTheorem 33, so there is no loss in generality.)

If there is a tiling, each white cell in the white patch has an asso-ciated black neighbor that lies in the same domino, and there mustbe at least as many of these black cells as white cells. However, thepatch is bad so this is not the case, a contradiction. Therefore notiling exist.

We finally have a criterion that can work for all tilings. However,the problem is that it can be difficult to find a bad patch, or to showthere aren’t any. For example, in a region R, the flow theorem iseasier to apply (although, of course, you could find a bad patch.)

That does not mean it is not useful: We will use this theorem a fewtimes to prove some other things about tilings, and it also gives us away the come up with untileable regions easily as the next theoremsshow.

An extension of a region R is a region P such that R is a subregionof P. Informally, an extension of R is some region formed by addingcells to R.

Theorem 35. Suppose R is a region with an odd number of cells. If we ap-ply the checkerboard coloring such thatW(R) < B(R), then any extensionof R that adds no new neighbors to black cells in the polyomino is untileable.

[Not referenced]

Proof. Let the original region be R, and its extension P. Since W < B,R is a bad patch with respect to itself. But since we do not add anyneighbors to form P, R is also a bad patch with respect to P, andtherefor untileable.

Theorem 36. If a region has a bad patch, it has a connected bad patch.

[Not referenced]

Proof. Let S be the bad patch. WLG assume that B(S) < W(S). Nowpartition S into connected disjoint sets Si such that the (black) neigh-bors of all the white cells in Si is also in Si. We want to show one ofthese sets is a bad patch. If each Si is a good patch, then B(Si) ≥

42 polyominoes

W(Si) for all i. But then B(s) = ∑i B(Si) ≥ ∑iW(Si) = W(S), acontradiction.

This theorem shows the process of generating untileable regionsdescribed above can generate every untileable region.

3.1.3 Cylinder Deletion

We can also use the Marriage Theorem (Theorem 34) to provecertain reductions do not affect the tileability of a region.

A vertical n-cylinder is a simply-connected region where each rowhas n cells (see Figure 37). A horizontal n-cylinder has n cells in eachcolumn (Hochberg, 2015).13

13 Cylinders also play an important rolein tiling extensions (see for exampleTheorem 123) and tilings of the infinitestrip.

Figure 37: Examples of vertical 3-cylinders.

An n-cylinder is tileable by dominoes if n is even, since each rowhas an obvious tiling by horizontally-placed dominoes. (a) (b)

Figure 38: Cylinder deletion. (a) Thecylinder S is the shape contained in theyellow dotted line, and R is the entireregion. (b) R S. Note the barriers thatcannot be crossed by dominoes.

Suppose a vertical cylinder S is a subregion of R such that it sharesis top and bottom borders with the border of R. If we remove S fromR, and move the two pieces together, we get the new region R S.We call this operation a deletion. An analogous definition can bemade for a horizontal cylinder S. After a deletion, we my be left witha region that has a barrier. For example, in Figure 38 we delete acylinder from the double-T region. This yields a new shape with twointernal barriers that cannot be crossed by dominoes. In particular, itmeans the two top cells are not neighbors of each other. Note that wecannot delete a horizontal cylinder from the reduced region becauseof the barriers.14 14 We will later see how this geometrical

operation is equivalent to simplifyingthe border word algebraically.

The operation is also very similar todeletions on rhombus tilings, also calledcontractions. In this context, the deletedshape is called a de Bruijn section. Seefor example Chavanon and Remila(2006) and Chavanon et al. (2003).

Theorem 37. Let S be an n-cylinder with n even. Then R is tileable, ifR S is tileable.


Proof. WLG, assume S is a vertical cylinder.Suppose R S is tileable. Then we can find a tiling for R as fol-

lows: In R S, the cut line is either covered by a domino or not.Tile all the cells in R that are also in R S the same way. If there aredominoes that cross the cut in R S, there will be two dominoes that

dominoes i 43

cross the two cuts in R (in the same row). So any row in S will eitherhave n untiled cells, or n− 2 untiled cells. Since this number is even,we can fill the row with horizontal dominoes. this gives us a tilingfor R.

Unfortunately, the converse of this theorem is not true (see for ex-ample Figure 39). However, a somewhat weaker version of it is true.A deletion is called safe when all the cells in R S have neighbors inthe same directions as they had in R.

Figure 39: An example of a region Rthat is tileable, but R S is not, for the2-cylinder S marked pink.

Theorem 38. Suppose S is a n-cylinder of R with n even, and that deletingit from R is safe. Then if R is tileable, then so is S R.15

15 I suspect something stronger is true,but have not been able to work out thedetails.


Proof. The reduced region R S has exactly the same neighbor setupas R. If R is tileable, then by the marriage theorem (Theorem 34),each set S of white cells has at least |S| black neighbors. This mustalso hold for R S, since if it had a bad patch, so would R. There-fore, R S is tileable.

This method gives us a way to reduce some regions to a moremanageable level.

But this method also gives us an algorithm to construct a tiling forR if we can find a tiling for R S.

Example 11. In Figure 40 we show how a tiling can be constructed for agiven region.

First (shown in the left column of Figure 40), we successively delete 2-cylinders from the region, until the resulting region is simple enough to tile.We then tile it (if we couldn’t, then its possible that no tiling exists for R.We do not know for sure, unless all the removals were safe).

Then, working backwards (shown on the right), we reinsert cylinders un-til we have rebuilt the original region. If the cutline goes between dominoes,we simply insert a domino perpendicular to the cutline; otherwise we inserta domino domino perpendicular to the cutline offset by a cell.

Problem 15. Establish the tileability of the regions in 27 by deleting cylin-ders until each region is manageable.

44 polyominoes

Figure 40: How to use cylinder deletionto find a tiling. On the left, from topto bottom, we delete 2-cylinders untilwe cannot. This region is easy to tile,shown on the bottom right. We thenreinsert 2-cylinders in the reverseorder (shown on the right-hand sidefrom bottom to top). At each stage wecomplete the tiling in the obvious way.

dominoes i 45

Figure 41: A bar graph with notationB(2 · 3 · 2 · 5 · 3 · 4).

Figure 42: A Young diagram withnotation B(5 · 3 · 22 · 1).

Problem 16. Give some examples of regions that do not allow us to delete acylinder from them. Are any of them tileable?

Problem 17. Can you characterize the regions from which we can deletecylinders?

A bar graph16 is a polyomino with columns all starting on the

16 Also called Manhatten polyomino (Seefor example Bodini and Lumbroso,2009). The notation used here is fromMartin (1986).

same horizontal line (Bousquet-Mélou et al. (1999)). A bar graph isuniquely identified by the number of cells in each column, and wewrite B(a1 · a2 · · · an) for a bar graph with ai cells in column i. We useexponentiation for repeated columns.

A Young diagram17 is a bar graph with columns in non-increasing

17 Also called a Ferrers diagram (eg.Delest and Fedou, 1993) or trapezoidalpolyomino (eg. Bodini and Lumbroso,2009) or partition polyomino (eg. Lerouxet al., 1998).

order (eg. Pak, 2000). In a Young diagram, we have ai ≥ aj wheni < j. 4

Theorem 39. A Young diagram of the form B(n · n− 1 · n− 2 · · · 2 · 1),with n > 0, is unbalanced (and therefor untileable).18

18 Compare this with Problem 10. Bothare consequences of Lemma 4 in Thiant(2003). That lemma requires the notionof a staircase that we don’t define here.


Proof. Apply the checkerboard coloring to the Young diagram suchthat the last column is black (Figure 43). Then

B−W =n

∑i=1

⌊i + 1

2

⌋−

n

∑i=1

⌊i2

⌋

=n+1

∑i=2

⌊i2

⌋−

n

∑i=1

⌊i2

⌋=

⌊n + 1

2

⌋−⌊

12

⌋=

⌊n + 1

2

⌋> 0.

Figure 43: A Young diagram of theform B(n · n − 1 · · · 1), in this caseB(5 · 4 · 3 · 2 · 1).

So the polyomino is unbalanced, and therefor untileable (Theorem23).

Theorem 40. A balanced Young diagram must either have at least twoadjacent columns equal, or at least two adjacent rows equal.


Proof. If a Young diagram does not satisfy those conditions, it mustbe a polyomino of the form B(n · n− 1 · n− 2 · · · 2 · 1), which is notbalanced by Theorem 39. Therefore, a balanced polyomino cannothave this form, and so a balanced polyomino must satisfy the condi-tions given.

Theorem 41. A Young diagram is tileable if and only if it is balanced19.

19 See also Problem 13.

46 polyominoes


Proof. If. ByTheorem 40 we know the polyomino has either two adjacent columnsof equal length, or two adjacent rows of equal length. In either case,we can delete a 2-cylinder from the region. Since the resulting regionmust still be balanced, the conditions apply again, and we can repeatthe process. The process must eventually end with the empty region,and this proves the Young diagram is tileable.

Only if. If a Young diagram is tileable it is balanced by Theorem23.

Figure 44: The relationship betweensome polyomino classes. Most of theseare obvious.

See also Problem 33 and Theorem53.

We already saw a class of polyominoes for which the Checker-board Criterion (Theorem 23) is enough (namely Young diagrams,see Theorem 41), and we may wonder if there are other classes ofpolyominoes this is true.

There is, and this section we prove that a large class of polyominoes—that include Young diagrams—are tileable when they satisfy thecheckerboard criterion. I will give three proofs of this fact, but beforewe get there we need a few definitions and helper theorems.

A domino in a tiling of some region is called exposed if at leastone of its long sides lies completely on the border of the tiled region.

Theorem 42 (Neretin (2017)). Any domino tiling with more than one tilehas at least two exposed dominoes.


Proof. We give an algorithm to produce two exposed long edges.

dominoes i 47

(1) Take any tile. If it is horizontal, look at its top (long) side.Otherwise look at the right (long) side.

(2) Introduce the coordinates: (x, y) = (0, 0) at the middle of thatside.

(3) If that side is free, we’re done. If it is not, there must be an-other tile blocking it (maybe partially). Switch to that tile (orthe rightmost/topmost of the two, if there are two).

(4) If it is horizontal, look at its top side. Otherwise look at theright side.

(5) Check the value of x + y at the middle of that side. Make sureit increases when we step from a horizontal tile to vertical orvice versa, or (in the worst case) stays constant when we stepto a tile of the same orientation.

(6) Go to step 3 and continue. It must end somewhere, for thereare only so many tiles and they never repeat. (There can’t bea cycle of tiles in different orientation, because x + y increaseswhen we change orientation, and never decreases. Neither canthere be a cycle made of horizontal tiles only, for in that case ysteadily increases in every step.)

To locate the second exposed long side, return to the initial tileand repeat everything in the opposite direction.

In fact, there must be at least two opposite long edges exposed.The procedure to find the first edge always produces a right or top

edge, and the procedure to find the second edge always produces aleft or bottom edge. If the edges are opposite, we are done. Supposethey are not different, and WLG let the first edge be a right edgeand the second edge a bottom edge. Use the procedure to find athird edge, this time going top left (rotate the coordinate system 90

degrees anticlockwise). The procedure must either produce a leftor top edge; in either case, it is opposite with one of the other twoexposed edges.

Figure 45: An example of a tiling thatachieves the minimum number ofexposed edges.

There are tilings that achieve the minimum number of exposededges. An example is shown in Figure 45.

It is easy to modify this proof to apply to all rectangles (in the caseof a square, all edges are “long”).

Problem 18. Extend Theorem 42 to all rectangles. Can it be extended to aneven bigger class of regions?

Rows are comparable if the column coordinates of one is a subsetof the others.

48 polyominoes

A polyomino in a fixed orientation is row convex when in eachrow, there are no spaces between any two cells.

Figure 46: A stack polyominocorresponding to the vectorB(1 · 2 · 2 · 3 · 5 · 4 · 2 · 1).

A stack polyomino is a bar graph that is row-convex and each pairof rows is comparable.20

20 Also called a trapeze Beauquier et al.(1995).

Theorem 43. A balanced stack polyomino contains a cylinder that can bedeleted.


Proof. If the top row has two or more cells, we can remove a verticalcylinder and we are done.

Suppose then the top row has only one cell, and that it is in col-umn k. Now we can remove a cylinder if for some i < k the columndiffers by the next by 2 or more and for some j ≥ k the column differsby more than two cells.

So suppose one of these conditions fail, WLG suppose for all i < kthe column differs from the next by just one cell (if they are equal, wecan remove a vertical cylinder).

Now divide the figure into two figures, with all m columns i ≤ kin S1 and the other n columns in S2. The deficiency in S1 is given

by ±⌈

m+12

⌉. If n < m, it must have a abslute deficiency that is less

then that of S1. If n > m, there must be two columns that are equal.And if n = m, we have two cells in the top row. Therefor, none of theconfigurations are possible, and therefor both conditions must hold,and therefor, we can remove a cylinder.

Note that if we know the region is tileable, we can use Theorem42 to prove a cylinder can be removed. However, knowing that wecan remove a cylinder from a tileable region is not as useful, sincewe often want to remove cylinders to establish whether a region istileable or not in the first place.

Theorem 44. Suppose R is a stack polyomino, and S is a cylinder that canbe deleted from R. Then R S is a stack polyomino.


Proof. Suppose the polyomino is B(a1 · · · am · · · an), and am−1 < am ≥am+1.

Suppose the cylinder we delete is vertical. This is equivalent toremoving two adjacent columns; the resulting polyomino is a bargraph, whose columns still satisfy thee inequalities that make thepolyomino a stack polyomino.

Suppose the cylinder we delete is horizontal, and it lies in columnsk to k′. Then, ak−1 ≤ ak + 2, and ak′ + 2 ≥ ak′+1. The new poly-omino is a bar graph with B(a1 · · · (ak − 2) · · · (am − 2) . . . ak′2 · · · an).

dominoes i 49

Combining all the inequalities, we see this new bar graph is a stackpolyomino.

Figure 47: A balanced stack polyomino.

A jig-saw region is a region formed by removing all cells of onecolor from the top of a balanced stack polyomino (Bougé and Cos-nard, 1992). A jig-saw region is sometimes, but not always, a stackpolyomino. It is always a bar graph.

Theorem 45 (Bougé and Cosnard (1992)). A balanced jig-saw region istileable by dominoes.


Proof. Suppose R is a balanced jig-saw region. Now make a partialtiling by placing a domino on each cell in the top row and its neigh-bor, and place horizontal dominoes on the second row starting fromthe outermost cells. If we remove all cells that are covered by domi-noes, the resulting region R′ is balanced, and a jig-saw region withone less row. Eventually, we must arrive at the empty figure, and soby the partition theorem (Theorem 2), R must be tileable.

Theorem 46 (Bougé and Cosnard (1992), Beauquier et al. (1995)). Abalanced stack polyomino is tileable.21

21 A fourth proof can be derrived froma more general theorem in Beauquieret al. (1995). Their result applies totiling a stack polyomino with two bars,one horizontal and the other vertical,with no rotations allowed. If both barshave length two, we have the case ofdominoes where any orientation isallowed. They introduce a lot of ma-chinery to prove the general theorem.However, if it is made specific for domi-noes, the ideas used resemble those ofthe second proof given here.


Proof 1. We can delete a 2-cylinder from the stack polyomino (The-orem 43) and the result a new stack polyomino (by Theorem 44) orthe empty region. We can continue this process until the last region isempty. By Theorem 37 this means the region is tileable.

Figure 48: The jig-saw polyomino thatcorresponds to the stack polyomino inFigure 47.

Proof 2. (Bougé and Cosnard (1992), Korn (2004, Part of Theorem11.1, p. 156)) There must be at least two sides with length greaterthan two (Theorem 42), and so at least one side that is not the base.If this side is vertical, remove the two cells connected to this side andthe top corner; if this side is horizontal, remove two cells connectedto this side and the left or right corner. In all cases, the remainingfigure is also a stack polyomino. Eventually, we must arrive at asingle domino. Re-inserting dominoes in the reversed order in thesame positions now yields a tiling of the stack polyomino.

Proof 3. (Bougé and Cosnard, 1992) Suppose the top row has an oddnumber of cells. Then put horizontal dominoes on the top row toleave one cell not tiled. If we remove all cells tiled, we get a newregion R′ which is a balanced jig-saw region, which is tileable (Theo-rem 45). Therefor, R must be tileable.

50 polyominoes

3.1.4 Coloring

We may wonder if other colorings exist that could give us informa-tion when the checkerboard coloring does not.

In a sense, the checkerboard coloring gives us the best informationwe can hope for in the general case. However, other colorings give usinformation in specific cases.

We first see what happens if we add a color.Let us color a region with three colors, amber, blue, and cherry,

such that colors in each row cycle, and diagonals have the same color.We will call this type of coloring a flag coloring.22 22 Golomb (1996, p. 4) calls this (jok-

ingly) a patriotic coloring, presumablybecause he used the three colors of theAmerican flag. We define generalizedflag colorings in Section 5.2.

When we place a domino, it always covers two different colors,and there are three such arrangements. Let’s denote the number oftimes a domino of each type is used by kAB, kAC, and kBC, and thenumber of cells of each color in a region by A, B, and C. We have thefollowing equations relating these values:

A = kAB + kAC (3.4)

B = kAB + kBC (3.5)

C = kAC + kBC (3.6)

With a bit of algebra, we get the following solutions to the equa-tions above:

kAB =A + B− C

2(3.7)

kAC =A− B + C

2(3.8)

kBC =−A + B + C

2(3.9)

These must all be whole numbers and non-negative (because,remember, they correspond to the numbers of dominoes), and thisbecomes a criterion: for a tiling to exist, it is necessary that kAB, kAC

and kBC are all non-negative integers.Figure 49 shows an example of a region that is balanced, but does

not satisfy this new criterion.

Figure 49: A region that is balanced, butdoes not satisfy another color criterion.

If you experiment a bit with this criterion, you will find that theregions are quite pathological and in general it is usually easy to findout they cannot be tiled without using the criterion. What is goingon?

From the equations, we can make the following observations: kxy

will be an integer unless the number of cells is odd. So this part is oflittle help (we already know the number of cells must be even fromTheorem 1). For kxy to be negative, one of the colors must exceed thesum of the other two. And regions that does this tend to be spikyand uninteresting.

dominoes i 51

What if we changed the pattern? We have not used any detail ofthe pattern, except that neighboring cells cannot have the same color.If we color with only this restriction, it’s rather easy to come up witha coloring where one color dominates. Figure 50 shows this coloringapplied to Figure 27(h).

Figure 50: A coloring showing theregion is not tileable.

But do we really need the two “loser” colors to be different? Theanswer is we don’t. Let’s set this up as we did before.

Color a region with amber and blue such that blue cells do nothave any blue neighbors. We call such a coloring a discriminatingcoloring, with blue cells the isolated color. The tiles can now be oftwo types, blue-and-amber and amber only, and let’s use kAB andkAA to denote how many of each tile we have.

Figure 51 shows this coloring applied to Figure 27(h).Figure 51: A coloring showing theregion is not tileable. There are 14 bluecells, and 12 yellow cells. Since bluecells have no neighbors, if the regionwas tileable we would have at least atmost as many blue cells as yellow cells.

We now have the following equations:

A = kAB + 2kAA (3.10)

B = kAB (3.11)

Solving, we get

kAB = B (3.12)

kAA =A− B

2(3.13)

(3.14)

For a tiling, kAB and kAA must be non-negative integers. kAB willalways be a non-negative integer; however, kAA will be negativewhen B > A. The number kAA will always be an integer as long asthe number of cells is even.

If a region has a coloring for which kAA is negative, we call theregion unfair by that coloring.

Example 12. Color the double-T as shown in Figure 12. With this coloring,B > W, and since no black cells have black neighbors, we conclude the tilingis impossible.

Figure 52: Double-T with alternativecoloring

Problem 19. For each remaining region in Figure 27, find a discriminatingcoloring by which it is unfair.

We state this as a theorem that we will call the two-color criterion.Note that it actually generalizes (and contains) the checkerboard cri-terion. However, the concept of balanced polyominoes is still useful,and when it works it’s a very efficient way to describe the particularcoloring.

Theorem 47 (Two-color criterion). If there is a discriminating coloring bywhich a region is unfair, then the region not tileable by dominoes.

52 polyominoes


Proof. WLG let black be the isolated color. If a tiling exists, theremust be b dominoes that cover a black and white cell, and w domi-noes that cover only white cells. The number of black cells in theregion is given by B = b, and then number of white cells W = 2w + b.Now if B > W, then b > 2w + b, and hence 2w < 0, which is impos-sible (the number of white only dominoes must be positive or zero).Therefore, no such tiling exists.

Problem 20. Show that we cannot get more information by adding morecolors using this type of criterion.

In a sense, the new color theorem is really just the marriage the-orem in disguise. The following theorem shows the connection be-tween them.

Theorem 48. A region has an unfair discriminating coloring if and only ifit has a bad patch.


Proof. If. Color the region with the checkerboard coloring.If the region has a bad patch, it has a black bad patch (Theorem 33).Let S be the biggest set that contains the black bad patch and is also abad black patch.

Then |W(S)| < |B(S)|, and so |W(R− S)| ≥ |B(R− S)| (if it wasnot, we could make S bigger). Now swap all the colors in R− S. Wenow have a coloring in which there are more white cells than blackcells, and all white cells only have black neighbors. Therefore wehave an unfair discriminating coloring.

Only if. Suppose we have an unfair discriminating coloring of aregion. Then the region is untilable (Theorem 47), and hence, in acheckerboard tiling, there is a bad patch (Theorem 34).

From this theorem, we get the following:

Theorem 49. A region is tileable if and only if all its discriminating color-ings are fair.


Proof. If. Suppose all the discriminating colorings of a region arefair, but it is untilable. Then, by Theorem 34 there is a bad patch,and by Theorem 48 there exists an unfair discriminating coloring,which contradicts our initial assumption, therefore, the region mustbe tileable. Only if. Suppose a region is tileable. If it had a unfair

dominoes i 53

discriminating coloring, it would not be tileable (by Theorem 47), andso all discriminating colorings must be fair.

At the beginning of this section, we said that the checkerboardcoloring gives us the best information in the general case. The reasonfor this is that of all discriminating colorings, the checkerboard hasthe highest density of isolated color, and can therefor discriminatethe most figures (of a certain area). I am not going to make this ideamore precise here. Furthermore, it makes ideas such as flow, and (aswe will see in another essay) height functions possible, which otherdiscriminating colors do not.

3.2 The Structure of Domino Tilings

Some tileable regions have cells that are tiled the same way inany tiling of the region. We call such cells frozen.

As we will see, there are three ways in which cells can becomefrozen:

• When it is part of a region that can geometrically only be tiledone way (when it is part of a peak, a notion we shall defineshortly).

• When other placements violate the border crossings theorem21.

• When other placements violate the flow theorem (Theorem25).

We will see that cells that are not frozen are always part of someclosed strip, and that by rotating the closed strip we can find a newtiling of the region.

A key point from this section is the following: any two tilings areconnected through a series of simple operations that involve rotationson closed strips of dominoes.

3.2.1 Closed strips

A path is a sequence of lattice points v1, v2, . . . , vn such that vi+1 andvi are neighbors, and no cell is repeated in the sequence.

Figure 53: Strip polyomino

A strip polyomino is a polyomino whose cells form a path. If wealso have vn neighbors v1, then we call the strip polyomino closed.23

23 In Beauquier et al. (1995) the wordring is used for a closed strip.

Figure 54: A closed strip

Theorem 50 (Beauquier et al. (1995)). If we apply a checkerboard coloringto a strip polyomino, and its two ends have opposite colors, the path has aneven number of cells, otherwise it has an odd number of cells.

54 polyominoes


Proof. We note that it is true for a strip polyomino with area 1. Nowsuppose it holds for strip polyominoes with area k. Consider now astrip polyomino with area k + 1, and suppose its two ends have thesame color. Now remove one end, to get a path of length k. Since thenew end must have opposite color of the one removed, the two endsare now of opposite color, and hence n must be even, and so k + 1must be odd. Suppose then the two ends have different colors. If weremove one end from the strip polyomino, the new polyomino musthave ends of the same color, and hence k must be odd, and so k + 1must be even. We proved that the theorem also holds for k + 1, andtherefore, it must hold for all k.

Theorem 51 (Beauquier et al. (1995)). Strip polyominoes with even areaare tileable. Closed strip polyominoes have even area, and each has eat leasttwo different tilings.

[Referenced on pages 54, 55, 56, 57, 59, 60, 100, 106, 117 and 119]

Proof. The cells of the strip polyomino forms a path v1, v2, . . . , vn,with n even. We can therefore partition the cells into sets {v1, v2},{v3, v4}, . . . , {vn−1, vn}, each containing two elements. Since in eachset the two cells are neighbors (Theorem 2), the set can be tiled by adomino, and so the entire region can be tiled by dominoes .

If the strip is closed, then vn and v1 are neighbors, and there-for has opposite colors, and so n must be even (Theorem 50). Onetiling is given above; another tiling is given by {vn, v1}, {v2, v3}, . . . ,{vn−2, vn−1}.

When trying to find a tiling of a region by hand, it is often easierto see if you can divide it into even strips (because you can quicklydraw lines through a region, and just check that endpoints havedifferent colors). This also helps you identify suitable choke points ifno tilings exist so that you can apply the flow theorem (Theorem 25)and prove it. The following two theorems illustrate this idea.

Theorem 52. All even-area rectangles are tileable.


Proof. If the area of a rectangle is even, then either its width or heightmust be even. WLG say the width is even. Now partition the rect-angle into rows. Each row is a strip polyomino with even area, andtherefore tileable (Theorem 51), and so is the whole region (Theorem2).

dominoes i 55

In fact, rectangles are strip polyominoes, and what is more, rectan-gles with even area are closed strips.

Theorem 53. Even-area rectangles are closed strips.


Proof. Let the width and height of the rectangle be m and n. WLGsuppose the width is even. Partition the rectangle into m + 1 strippolyominoes as follows: form the first m snakes Si from all cells ineach column except the top one; form the last snake Sm+1 from thecells in the top rows. To each strip assign a head and tail: Except forthe last snake, odd numbered snakes get a tail at the top and head atthe bottom. Even-numbered snakes get a head at the bottom and atail at the top. The last snake gets a head at the left and a tail at theright.

Now if we join the snakes head to tail in order, and join the headof the last to the tail of the first, we have a path through all the cellsin the rectangle, which proves it is a closed strip.

Figure 55: A rectangle is a strip poly-omino.

Call a strip polyomino with odd area and with black endpointsblack, and one with white endpoints white.

Theorem 54. Let R be a white strip polyomino.

(1) If we remove a white cell from the strip, the remaining figure(s) aretileable.

(2) If we add a black cell (that is not already part of R) to the polyomino,then the resulting figure is tileable.

The theorem holds of we interchange black and white.

[Not referenced]

Proof. If we remove a white endpoint, the resulting figure is a stripwith even area and thus tileable (Theorem 51). If, on the other hand,we remove another white strip vi, then v0, · · · , vi−1 and vi + 1, · · · , vn−1

are two strip polyominoes, both with even area. (This is because v0 iswhite, and vi−1, a neighbor of vi, is black, and so Theorem 50 applies;the same for the other strip.)

If we add the black cell to one of the strip polyominoes endpoints,we have a strip polyomino with even area, and so it is tileable. If, onthe other hand, we add a black cell u to a strip polyomino of oddarea next to a cell that is not an endpoint, vi (necessarily white), thenwe can partition the figure into three figures: {u, vi}, {v0, · · · , vi−1}and {vi + 1, · · · , vn−1}. The first is a domino, and hence tileable. Thelast two are strips with even area (as argued above) and hence aretileable.

56 polyominoes

It is easy to extend this theorem to remove a white strip or add ablack strip (that don’t overlap with the original).

Let R be a region, and let R1 be all the cells on the border of R. LetSi = R − Ri, and Ri are all the cells on the border of Si. We call Rsaturnian if for some k,

(1) Ri = ∅ for all i > k,

(2) Rk is tileable by dominoes

(3) Ri is a set of closed strips for all i < k.

Si need not be connected. If a cell c is in Ri, we define its level bylev(c) = i. (Parlier and Zappa, 2017)

The number k is called the number of levels of R; and the subre-gion Rk is called the core.

Figure 56: Examples of Saturnianregions. The third figure is an exampleof a Saturnian figure for which R2 isdisconnected.

Theorem 55. Saturnian polyominoes are tilable by dominoes.


Proof. We can partition R as follows R = R1 + R2 + · · · Rk. Each ofRi<k is a closed strip and tileable (Theorem 51), and Rk is tileable bydefinition; hence, so is R (Theorem 2).

Problem 21.

(1) Prove that rectangles with even area are Saturnian.

(2) When is a rectangle with some corners removed Saturnian?

(3) Prove that the core of a Saturnian region cannot contain a 3 × 3square subregion.

A peak is a cell of a region with only one neighbor in the region(Beauquier et al., 1995).

Figure 57: Snake

dominoes i 57

A snake is a polyomino that has two peaks, and every other cellhas exactly two neighbors (Goupil et al., 2013). Snakes are strip poly-ominoes, and therefore snakes with even area are tileable.

A ring is a polyomino where each cell has exactly two neighbors.Rings are closed strip polyominoes.

Rings are closed strips, and therefore they have even area, andexactly two tilings.

Although we prove theorems for strip polynomials, it is not ingeneral easy to determine whether a given polyomino is a strip poly-omino (for example, it is not immediately obvious that Figure 27(e)is not a strip polyomino). On the other hand, it is easy to recognizesnakes and rings.

k P(k) S(k)A000105 A002013

1 1 1

2 1 1

3 2 1

4 5 2

5 12 3

6 35 7

7 108 13

8 369 31

9 1285 65

10 4655 154

α 3.6 2.3

Table 6: Number of free snakes S(k)compared to the number if free poly-ominoes P(k).

Problem 22. Define the dilation of a region as the region with all cells ofthe original region and their neighbors. Let R′ be the dilation of R. When isR′ − R a ring?

Figure 58: Ring

Problem 23. Define the erosion of a region as the region with all cells of theoriginal region that has four neighbors. Let R′ be the erosion of R. When isR′ − R a ring?

Problem 24. What if in the definitions in the above we use 8-neighborsinstead of 4-neighbors?

Theorem 56 (Gomery’s Theorem). If we remove a white and black cellfrom a checkerboard colored strip polyomino, the remaining region is tileableby dominoes.24 24 The theorem, by Ralph Gomery, was

originally given in only for the 8× 8square (Honsberger, 1973, p. 65–67),and the proof was slightly different.However, the main idea of the proof isthe same as given here.


Proof. Let the cells along the path of the strip polyomino be v1, v2, v3,. . . , vk, and suppose the removed cells are vm and vn with m < n.

If the removed cells are consecutive in the path (Figure 59), thenthe region has a path vn+1, vn+2, . . . vk, v1, v2, . . . , vn−2. This is an evenstrip, and so is tileable (Theorem 51).

Figure 59: Removing two cells consecu-tive in the path.

If the removed cells are not consecutive in the path (Figure 60),they partition the strip into two strips. The ends of each strip neigh-bor cells of opposite color, and therefore the ends of each strip is ofopposite color. Therefore, the strips are tileable, and so is the wholeregion.

Figure 60: Removing two cells notconsecutive in the path.

3.2.2 Frozen Cells



58 polyominoes

In Figure 62, we can see a domino can fit only one way in the twoyellow cells. We can then remove these cells, and consider whetherthe rest of the region can be tiled. Since this is a 2× 2 square, it canbe, and so we conclude the whole region is tileable. The same tech-nique works for more complicated regions, such as in Figure 61.

Figure 61: Illustrating peak removal.

When you apply this technique to Figure 27(b), you get a discon-nected monomino after the first step, which is not tileable, and so thewhole region is not tileable.

We will now formalize this process in a series of definitions andtheorems.

A subregion of R is frozen if it can be covered by dominoes inonly one way in any tiling of R.

Figure 62: The yellow cells can only becovered in one way by a domino.

Theorem 57. Suppose a region R is partitioned into two partitions, S1

and S2, and S1 is tileable, and would be frozen in any tiling of R if it exists.Then the region is tileable if and only if S2 is tileable.


Proof. If. If S2 is tileable, then R is tileable by Theorem 2.Only if. Suppose R is tileable. Because dominoes can cover the

frozen partition in only one way, in a tiling of R the dominoes thatare part of S1 must also tile S1, since it is indeed tileable. Therefore,the dominoes not part of S1, must tile S2.

Figure 63: A region with 4 peaks,shown in yellow.

Theorem 58. In a tileable region, a peak and its neighbor are frozen.

[Not referenced]

Proof. There is only one way for the domino to cover the peak, and inthat one way it also covers the neighbor. So if the region has a tiling,the neighbor cell can also only be covered one way.

Figure 64: A region whose compactsubregion is empty.

From this theorem, it follows that a region with a peak is tileable ifand only if the region(s) that remain after we remove a peak and itsconnected cell. We can repeatedly remove peaks and their connected

dominoes i 59

cells until no peaks remain. The resulting region (which may not nec-essarily be connected) is called the compact subregion. The compactsubregion of R is denoted R∗. Note that R∗ can be empty (Figure 64),or disconnected (Figure 65). It is possible that R∗ = R, in which casewe call R compact.

Figure 65: A region whose compactsubregion (blue) is disconnected.

Problem 25. Suppose R has no holes. Prove R∗ does not have any holes.

Problem 26. What is the bound on ∆(R) as a function of the number ofcells in R if R is compact?

Theorem 59. A region R is tileable if and only if its compact subregionR∗ is tileable.

[Not referenced]

Proof. Let R′ = R− R∗, so that R′ and R∗ are partitions of R. Since allthe cells in R′ are frozen, by Theorem 57 R is tileable if and only if R∗

is tileable.

If a region consists out of disconnected parts, it is tileable if andonly if each part is tileable (Theorem 2.) The tileability of all regionsthus boils down to whether connected compact shapes are tileable.

Theorem 60 (Beauquier et al. (1995), Theorem 3.5). If a region has aunique tiling, it must have at least two peaks.


Proof. Suppose a region has no peaks, and that it has a tiling. Nowconstruct a sequence of cells as follows: v1 is any cell, and v2 is thecell in the same domino. Choose v3, a neighbor of v2 that is not v1

(we can do this, since v2 is not a peak).We can continue in this fashion, always choosing new cells that

haven’t been chosen before, until eventually, we are “trapped” orrun out of cells. Now the last cell, vn, must have a neighbor that isnot vn−1, and is already part of the sequence, say vi. (If it did not,then vn would be a peak). So we have a sub-sequence vi, vi+1, . . . vn

that is a closed strip. Closed strips are tileable in more than one way(Theorem 51), and therefore, the entire region must be tileable inmore than one way.

Suppose the region does have one peak. Construct the same se-quence as above, but choose v1 as the peak.

This implies that compact shapes have at least two tilings. Figure66 shows a big figure with a unique tiling, mentioned in Pachter andKim (1998).

60 polyominoes

Figure 66: A figure with a unique tiling

If we recursively remove peaks and their connected cells, we areeventually left with either a compact region, or the empty region. Ifthe former is the case, the region is uniquely tileable. Otherwise, ithas more than one tiling.

3.2.3 Tiling Transformations

In this section we look at how one tiling can be transformedinto another with a series of “primitive” transformations.

Theorem 61. A cell is not frozen if and only if it is part of a tiled closedstrip.


Proof. If. Suppose a cell v1 is part of a tiled closed strip, and itsneighbor in the the same domino is v0, and its other neighbor alongthe strip is v1. Then by Theorem 51 there is another tiling of the strip,and in that tiling v1 and v2 are in the same domino. Therefore, thereis more than one way for v1 to be covered by dominoes, and thereforeit is not frozen.

Only if. Suppose a cell v1 can be tiled in two ways. In one tiling v0

is the same domino, and in the other tiling v2 is in the same domino.Then let v3 be v2’s neighbor in the first tiling, v4 is v3’s neighbor in

dominoes i 61

the second tiling, and so on. We can always extend the sequence, butsince there are only a finite number of cells in the region, eventuallythe next cell must be one that is already in the sequence. Moreover,this repeated must be v0, since a cell can be in the same dominoin two different tilings at most two times, and any other cells vi isalready in the same domino as either vi−1 and vi+1.

Therefore we have a sequence of cells v0, v1, . . . , vk, where vi+1

is a neighbor of vi, and vn is a neighbor of v0. Therefore, the cellsv0, v1, . . . , vn form a closed strip.

Figure 67: An example of a closed stripthat contains a point and is tiled in bothtilings.

The regions in Figure 68 all have frozen cells that cannot be part ofany tiled closed strip.

Problem 27. Show that if we insert a cylinder into a figure where all thecells that surround the insertion section is frozen, then frozen cells stayfrozen, and the entire cylinder is also frozen.

From the result of the problem above, we know that we can getthe ratio of frozen cells to total cells as close to 1 as we want, even infigures without peaks, bridges, or holes.

(a) (b) (c) (d)

(e) (f) (g)

(h) Figure 68: Examples of regions withfrozen cells, marked with yellowstripes.Theorem 62. If two closed strips are put side-to-side, then together they

form a strip polyomino.

62 polyominoes

[Not referenced]

Proof. Let the two closed strips be u1, u2, . . . , um, and b1, b2, . . . , bn,and suppose ui is neighbors with bj. Then a strip is given by

ui+1, ui+2, . . . um, u0, u1, . . . , ui, vj, vj+1, . . . , vn, v0, v1, . . . vj−1.

Figure 69 shows an example of 3 connected closed strips (2 × 2squares) that are connected, but do not form a strip.

Figure 69: An example of 3 connectedclosed strips that do not form a strip.

An analogous argument shows any non-frozen cell tiled by a ver-tical domino, if it has a non-frozen horizontal neighbor, it can also betiled horizontally.

In a region, if we replace a closed strip with its dual tiling, we callthis operation a strip rotation25 (Figure 70). When the closed strip is 25 In Propp (2002)the other calls this

operation (in the context of orientedgraphs), a face twist.

a 2× 2 square, we call the strip rotation a flip, and the two dominoesthat form the strip a flippable pair. In Theorem 61 we saw how toconstruct a closed strip that contains a cell from two different tilings.In the theorem below, we exploit this idea to prove that one tilingcan be transform into any other tiling through a sequence of striprotations.

Figure 70: An example of a strip rota-tion.

Theorem 63 (The tiling connection theorem). We can obtain one tilingfrom another by a sequence of non-overlapping strip rotations.


Proof. Pick any cell with a different tiling in the two tilings. Con-struct the closed strip as in Theorem 61, and perform a strip rotationon that strip in the first tiling. All the dominoes in that strip now hasthe same tiling as in the second tiling. Repeat the process. Notice,that subsequent rounds do not change any dominoes that are alreadyin the same position as in the second tiling, so non of the strips over-lap. Since the region is finite, and we reduce the number of dominoesthat differ from the second tiling in each step, the process must endwhen all the dominoes match.

Theorem 64. Let R be a tiled region, and S a subregion of R. Performing astrip rotation on R does not affect the flow on S.26 26 These ideas are more or less pre-

sented in Saldanha et al. (1995). Theyconsider flow through cuts that do notdisconnect the region.[Not referenced]

Proof. This is clearly true when the closed strip is entirely within S orentirely out of S.

Suppose then the closed strip goes over the border of S.If a strip outside S meets S at cells u and v, then either the strip

has an even number of cells (and so u and v must have opposite

dominoes i 63

colors, by Theorem 50), or an odd number of cells (and u and v musthave the same colors, by Theorem 50).

In the first case, dominoes must either cross S at both points, orneither, and the net contribution of dominoes at those two pointsare zero. Doing a strip rotation will cause the opposite; either domi-noes don’t cross at either point, or they do at both. Again, the netcontribution to the flow is 0.

In the second case, a domino must cross at the one point, andnot the other, and therefore the contribution to the flow is +1 or−1, depending on the color of the cell where the domino crossesinside S. Say the crossing happens at u, and that u is black. Then thedomino contributes +1 to the flow. v must be white too. When wedo a rotation, there is no crossing at u, and a crossing at v. Since vis white, the contribution to the flow is 1. A similar argument showswhen u is white the contribution before and after the strip rotation is-1.

So in all cases, for every strip, part of the closed strip, that meats S,we have that their contributions to the flow stays unchanged by striprotations, and so the overall flow is unchanged.

Theorem 65. Every tileable region contains at least one of the following:

• Two peaks.• A 2× 2 square subregion.• A hole.


Proof. Suppose the region has a unique tiling. Then it contains twopeaks by Theorem 60. Suppose then the region has more than onetiling. Pick a cell v that is not frozen. This cell must be part of aclosed strip. Inside this closed strip, choose the left-most bottom-most cell. Because it is part of a closed strip, it must have at twoneighbors inside the strip, and since this is a left-most bottom-mostcell, it must have a top neighbor vT and right neighbor vR. Now con-sider the position u to the top of vR. Either there is a cell, or there isnot. In the former case, we have a 2× 2 square. Suppose then thereis not a cell in that spot. Since all the cells in the closed strip can-not lie to the left of v, they must surround u, and therefore there is ahole.

The following theorem allows us to do induction on closed strips:it gives us a way of breaking closed strips into smaller closed strips.

Theorem 66. Every closed strip without holes and more than 4 cells con-tains a smaller tiled closed strip as subregion.

64 polyominoes


Proof. By Theorem 65 a closed script must have a either a hole or a2× 2 square as subregion, and since this closed script does not have ahole, it must contain a 2× 2 subregion.

Now consider how that square is tiled. Since the flux is 0, we musthave that 0, 2, or 4 dominoes overlap the border.

• If 0 dominoes overlap the border, we have a flippable pair,which is a tiled closed strip, and a subregion smaller than theoriginal.

• If 2 dominoes overlap, the cells where dominoes overlap mustbe neighbors. There are four ways in which we can form aclosed strip with this configuration (Figure 71). In two of themthe branches of the strip must overlap (Figure 71(a) and (b)),and in one of the remaining the two branches are untileable(Figure 71(c)). That leaves only one configuration (Figure71(d)).

(a)(b)

(c)(d)

Figure 71: Four ways in which closedstrips can be formed when 2 dominoesoverlap.In this configuration we can form a smaller closed strip by

removing the domino that lies completely within the square(Figure 72).

Figure 72: How a smaller closed stripcan be formed this configuration.

• If 4 dominoes overlap, there are several possibilities of forminga closed strip, some are shown in Figure 73. Most of thesehave branches that cannot be tiled, (Figure 73(a)), or leave outone of the dominoes (Figure 73(b)); many also have branchesthat intersect. Two cases work (Figure 73(c) and (d)).

(a) (b)(c) (d)

Figure 73: Some ways in which closedstrips can be formed when 4 dominoesoverlap.In these cases we can split the path into two as shown in Fig-

ure 74. Either path is shorter than the original.

dominoes i 65

(a) (b)

Figure 74: Splitting paths.

Theorem 67. A tiling of a closed strip without holes must have at least oneflippable pair.


Proof. If the closed strip is a 2× 2 square, the closed strip is a flip-pable pair and we are done.

Otherwise, by Theorem 66, the region contains a smaller closedstrip as subregion. Continue to find a smaller and smaller subregion.Since the tiling is finite, eventually we must find a 2 × 2 square,which is a flippable pair.

Theorem 68. A strip rotation (of a strip without holes) is equivalent to asequence of flips with the same orientation.


Proof. It is true for a closed strip of 4 cells.Find a flippable pair (it must exist by Theorem 67). If it is a 2× 2

square, a flip rotates the entire strip.Otherwise, the closed strip has more than 4 cells, and it can have

either one branch (Figure 75(a)) or two branches (Figure 76(a)) tocomplete the strip.

(1) If it has one branch. Do the flip (Figure 75(a)). One domino isnow in the right position, the other, with the branch, makes aclosed strip with less than n cells, so we can do a rotation byinduction. After this rotation, the complete strip is rotated.

(a) (b)

Figure 75: A strip with one branchbefore and after the flip.

(2) If it has two branches. Do the flip. We can now form twoclosed strips with one domino in each (Figure 76(a)). Each ofthese can be rotated by induction. After the rotation of each,the full strip is rotated too.

66 polyominoes

(a) (b)

Figure 76: A strip with two branchesbefore and after the flip.

Theorem 69. If a closed strip has a filled interior, a rotation of the strip isequivalent to a sequence of flips.


Proof. Find a domino with its long edge on the border of the interior(such a domino exist by Theorem 42). The long edge on the borderneighbors the outer closed strip in two cells; either they lie in thesame domino, or they don’t.

(1) If they don’t, they lie in two dominoes, who may or maynot be neighbors in the strip. If they are, we can form a newclosed strip by simply including the interior domino betweenthem. Otherwise, we form two strips, a closed one with theinterior domino between them, and the other what used to bebetween the two outer dominoes. These are closed if there ismore than one domino.

(2) If they do, we flip the pair, and loop them in.

We continue this process (in each case using either of the availableclosed strips constructed so far) until the entire interior is enclosedwithin one of the closed strips R1, R2, . . . . We may have sheddedseveral strips S1, S2, . . . in the process.

Now each of Ri is a closed strip without holes, so a rotation onthese are equivalent to flips. Perform the rotation on each.

Each of Si may either be a closed strip without holes, or a singledomino.

(1) In the case of the former, a rotation is equivalent to a sequenceof flips, so we can do a rotation.

(2) In the case of the latter, after all the rotations, either one of thedominoes that used to be a neighbor inside the strip is now aflippable pair with it, or it is not.

(a) If it is, do the flip.

(b) Otherwise, perform a sequence of flips as shown in Figure77.

dominoes i 67

Figure 77: A sequence of flips to rotatea shedded single domino with nowneighbors after the rotation.

All the dominoes in in the entire closed-strip are now rotated.Some dominoes in the interior might be in wrong positions, so theyneed to be restored. Pick a cell, and find a closed strip from the cur-rent and desired tiling. If this strip has an empty interior, we canperform a rotation equivalent to flips. Otherwise, we re-apply thistheorem (it must end, because with each re-application we have fewertiles and there is only a finite amount.) We repeat until the interior iscompletely restored.

We have performed a rotation of the outer strip, using only flipsand flip-equivalent rotations. Therefore, the entire operation is flip-equivalent.

Theorem 70 (Saldanha et al. (1995)). We can obtain one tiling of a regionwithout holes from another by a sequence of flips.


Proof. Any rotation of a strip is equivalent to flips, whether it’s in-terior is empty (Theorem 68) or filled (Theorem 69), and any tilingcan be transformed into another by a series of rotations (Theorem 63).Therefore, any tiling can be transformed into another with a series offlips.27 27 See also Rémila (2004) which gives

an easier treatment of flips and dominotilings.

Theorem 71. A tileable stack polyomino with the bottom two rows equalhas at least one flippable pair.

[Not referenced]

Proof. A stack polyomino can have at most 3 peaks (the first andlast cell in the bottom row, and the top row). So if the bottom rowshave the same number of cells, there can be a most one peak. But fortilings to be unique, we need at least two peaks (Theorem 60). So thetiling is not unique, and there must be another tiling. We can reachthe other tiling by a sequence if flips (Theorem 70), which means theoriginal tiling must have at least one flippable pair.

3.3 Further Reading

68 polyominoes

In Mendelsohn (2004) the author gives a gentle introduction to therelationship between domino tilings and graph theory.

We use a variety of color arguments elsewhere in these essays (forexample Sections 6.3 and 6.1.2). See Engel (1998) for other applica-tions and ways to structure coloring arguments.

As mentioned in a side note, the marriage theorem we presentedhere is a specific example of the much more general theorem in thearea of matching theory. For a survey on the development of matchinghistory, see Plummer (1992), and for a detailed treatment, see thebook Lovász and Plummer (2009). Many books on combinatorics andgraph theory contain chapters on matching, see for example Harriset al. (2008), Diestel (2000) and Bondy and Murty (1976).

We showed that multiple tilings in a region is the result of certainclosed strips (Theorems 60, 61). If we remove dominoes to removeall these closed strips, the tiling is unique. The number of dominoesto remove is called the forcing number (for example, if we remove ndominoes along the diagonal of a 2n× 2n square as in Figure 66, theresulting figure has a unique tiling. These ideas are discussed in forexample Pachter and Kim (1998) and Lam and Pachter (2003).

Related to the idea of frozen cells is the following: There are cer-tain edges in a region that will never be covered by a domino in anytiling. If we consider the dual graph, the edge is called a fixed sin-gle edge. (In this setup, there is an edge between two cells that areneighbors. A tiling is a matching of the graph, and edges can bedouble or single, depending on whether there is a matching edge ornot. Frozen tiles correspond to fixed double edges.) See for exampleZhang (1996). The edges (in the original region) that are not crossedby dominoes in any tiling are called fracture edges by Fournier (1997).

Domino tilings and their statistics are of interest to physicists be-cause they model the behavior of certain types of molecules. In thiscontext, a domino is usually called a dimer, a monomino is called amonomer, and the tilings may be considered on more general graphsthan square grids. The statistical model that deal with these tilings(or coverings) is called the dimer model. For a survey on the dimermodel, see Kenyon (2000c). In the paper Cohn et al. (2001) the au-thors give a detailed analysis of the statistics of arbitrary regions.

For a less naive view on the topics discuss here, I recommendthe following sequence of papers to get familiar with the algebra ofpolyomino tilings:

(1) Propp (1997) is a gentle introduction to Conway and Lagarias’swork,

(2) Hitchman (2017) expands on these ideas, and includes a dis-cussion on tiling invariants ,

dominoes i 69

(3) Conway and Lagarias (1990) is the paper where Conwayand Lagarias introduces some group theoretic tools to studytilings, and

(4) Thurston (1990) is where height functions are introduced forthe first time, and a polynomial time algorithm is described.

The thesis Donaldson (1996) discusses these ideas in a moreleisurely fashion, and among other things, works through a proofof Thurston’s algorithm.

For domino tilings, specifically, the following papers are a usefulstart28: 28 I will give a more expansive bibli-

ography after we covered more topics.The papers listed here deal more or lesswith the same topics than this essay

• Tiling figures of the plane with two bars (Beauquier et al., 1995).

• Spaces of domino tilings (Saldanha et al., 1995).

• The lattice structure of the set of domino tilings of a polygon (Rémila,2004).

• Optimal partial tiling of Manhattan polyominoes (Bodini andLumbroso, 2009).

4Dominoes II

In this essay, we look further into the structure of domino tilings,how to count domino tilings, and what happens if we add or removeconstraints from the tiling problem.

4.1 The Structure of Domino Tilings

We have seen that the domino tilings of a region are connectedthrough strip rotations, and indeed that if the region is simply-connected, that it is connected through flips .

In this section we examine this structure further. We are workingtowards ways to answer questions like the following:

• What is the minimum number of flips we must perform on atiling T to produce a tiling U?

• What can we say about how many (and where) flippable pairsoccur in a tiling of a region?

We will do this by showing that the set of tilings of a region have acertain structure–they form a distributive lattice.

4.1.1 Height Functions

In this section we develop the height function, which will allow usto define an important tiling criterion and algorithm. We alreadyencountered the two main ideas on which the height function isbased:

(1) We can know what is the deficiency of a region by only look-ing at the edges on the border. (This is Theorem 27).

(2) If a region is partitioned into two subregions by a cut, thelength of the cut must just about be double the absolute valueof the deficiency of the one region so that enough dominoescan overlap it to compensate for the deficiency (Theorem 26).

dominoes ii 71

The general idea is to look at a specific set of partitions, and seeif the cut length is always long enough among these. (Instead ofstraight cuts, we will use cuts with a different shape.) If it is not, forone of these partitions, then we know a tiling is impossible. If it islong enough for all these partitions, then we can conclude a tiling ispossible.

Suppose u and v are neighboring vertices. Then we define thespin sp(v, u) as 1 when we go from u to v and the cell to the left isblack, and −1 if it is white (Rémila, 2004). Suppose v0, v1, · · · vk isa path of vertices. Let us define the height difference of this pathh(v0, v1, · · · vk) = ∑k

i=1 sp(vk, vk−1) (Rémila, 2004)1. 1 Ito (1996) calls this the valueof the path, and use the symbol〈v0, v1, · · · , vk〉 to denote it.Theorem 72. If c0, · · · , cn are the vertices of the border of a closed region

(with c0 = cn ), then if the region is balanced, h(v0, v1, · · · vk) = 0.2 2 This is a stronger version of Rémila(2004, Proposition 1).


Proof. Along the border of a balanced region, deficiency is 0 (The-orem 23), and the difference in white and black edges are equalto 4 times the deficiency (Theorem 27), and therefor also 0, and soh(c0, · · · , cn) = ∑k

i=1 sp(ck, ck−1) = 0.

A path that does not cross any dominoes is called a domino edgepath Ito (1996).

Theorem 73 (Rémila (2004), Corollary 1). Suppose we have a tiling of aregion, and P and P′ are two different domino edge paths from vertex u tovertex v. Then h(P) = h(P′).

[Not referenced]

Proof. WLG assume the paths do not cross. (If the paths cross at avertex w, we can consider the two paths from u to w and from w tov.)

Let P = p0 · · · pm and P′ = p′0 · · · p′n (with p0 = p0 = u andp′m = p′n). The path Q = p0 · · · pm p′k−1 · · · p

′0 does not cross any

dominoes, its interior is therefore tileable, and therefor balanced(Theorem 23). So h(Q) = 0 (Theorem 72). But h(Q) = h(P)− h(P′),and so h(P) = h(P′).

This shows h only depends on the start and end nodes. We nowpick any vertex, v∗, and define h(v∗) = 0, and h(v) = h(v∗, v1, v2, · · · v)for any domino edge path. The function h is called a height function(Rémila, 2004, Definition 1).

The height function on the border of a region is the same for alltilings of a region.

72 polyominoes

Theorem 74 (Rémila (2004), Proposition 2). Given the height functionof a tiling, we can reconstruct the tiling. Therefor, if for two tilings we havehT(v) = hU(v) for all v, then T = U.


Proof. The height difference between neighboring vertices can onlybe:

(1) ±1, if there is not a domino that crosses the edge betweenthem, or

(2) ±3, if there is a domino that crosses the edge between them.

So for any two neighboring vertices we can determine from theheight function whether a domino crosses the edge between themor not, and doing this for all neighboring pairs will show us whereall the dominoes in the tiling lie.

Theorem 75 (Rémila (2004), Proposition 3). An integer function fdefined on the vertices of a region that satisfy the following properties is theheight function of a tiling:

(1) There exists a vertex v∗ such f (v∗) = 0.

(2) For neighboring vertices u and v such that sp(u, v) = 1, we eitherhave f (v) = f (u) + 1, or f (v) = f (u)− 3.

(3) If these vertices are on the boundary, then f (v) = f (u) + 1.


Proof. Let w0, w1, w2, w3, w4 = w0 be a path around a black cell(going anticlockwise). Then by (2) the only possibility is if there isone vertex wj such that f (wj+1) = f (wj)− 3, and for other the otherthree vertices we have f (wi+1) = f (wi) + 1 (otherwise, we cannothave f (w0) = f (w4)).

A symmetric argument can be made for white cells. We can nowplace a domino over the edges wi − wj+1. Note that dominoes can-not overlap, since if they overlapped in a black cell, for example, itmeans there must be two edges wj − wj+1, and not just one, such thatf (wi+1) = f (wi) + 1.

This gives us a tiling of the region. We can now verify that f (v) =h(v) by induction on the distance of v from v∗, which completes theproof.

Theorem 76. Let R be a region and S be a strip tiled in a tiling T of R.Let U be T with S rotated clockwise. Then the height of all vertices on andoutside the outside border of S stays the same. The heights of all vertices onand inside the inside border increases by 4.

dominoes ii 73


Proof. All vertices outside and on the outside border of the closedstrip can be reached by paths that are unaffected by the strip rotation;so these vertices’ height stays the same.

Now consider a vertex u outside the closed strip or on the outerborder, and one v on the inside (or on the inside border). Beforethe strip rotation, there is a path from u to v from which we cancalculate the height of v given that of u. This path must intersect thestrip at least once, and WLG lets say it does cross it once. Since itcannot pass through a domino, it must pass between two dominoeson the closed strip. Let the vertices where this happens be wo (on theoutside border) and wi (on the inside border).

After the strip rotation, we cannot go from wo directly to wi asbefore; instead, we need to pass around the domino. Suppose wego clockwise. Since the strip rotation was also clockwise, it meanswe are passing a black node on the left. We are also skipping theoriginal section with a white domino on the left. Therefor, the netdifference in height is 3− (−1), which means the new height of v isnow h′ = h + 4.

We will call a strip rotation up if it takes a anti-clockwise tiling toa clockwise tiling, and down otherwise.

A special case of this is when S is a flippable pair. An up flipincreases a single vertex’s height by 4. The height of a vertex betweentwo tilings can only differ by multiples of 4, because we can get onetiling from any tiling by strip rotations3. (Note that while these strip 3 This is Rémila (2004, Lemma 1).

rotations don’t overlap, one can completely surround another, and soit is possible for the height of a vertex to differ by more than 4.)

Problem 28.

(1) Suppose u and v are the two vertices of a cell edge of a tileable regionR. Show that for any two tilings T and U of R, that the value of(hT(v)− hU(v))− (hT(u)− hU(u)) is 4, 0 or −4.

(2) When is it true that if we have a tiling T of a region R, and a func-tion g defined on the vertices of R such that for any two vertices uand v that share a cell edge, we have (hT(v) − g(v)) − (hT(u) −g(u)) is 4, 0 or −4, then there is a tiling U such that hU = g.

Suppose T and U are two tilings of a region, with height functionshT and hU . Then we write T ≤ U iff hT(vi) ≤ hU(vi) for all i.

A relation ≤ defined on a set is called a partial order if it satisfiesthese three properties for any elements A, B, and C of the set:

74 polyominoes

A ≤ A Reflexive (1)

A ≤ B and B ≤ A implies A = B Commutative (2)

A ≤ B and B ≤ C implies A ≤ C Anti-symmetric (3)

Theorem 77. The operation ≤ is a partial order.


Proof. We need to prove the above three properties hold.

(1) hA(v) ≤ hA(v), and so A ≤ A.

(2) A ≤ B implies hA(v) ≤ hB(v), and B ≤ A implies hB(v) ≤hA(v), so hA(v) = hB(v). And therefor, A = B (Theorem 74).

(3) A ≤ B implies hA(v) ≤ hB(v) and B ≤ C implies hB(v) ≤hC(v), so we have hA(v) ≤ hB(v), and so A ≤ C.

Theorem 78. Suppose that T and U are tilings that differ by a flip, andT ≤ U. Then if X is a tiling such that T ≤ X ≤ U, then either X = T orX = U.


Proof. From Theorem 76 it follows that for any two tilings T and U,we have hT(v)− hU(v) = 4k for some integer k.

Now if T and U are two tilings that differ by a flip, then by The-orem 76 there is a single vertex u such that hT(u) + 4 = hU(u),and for all other vertices v we have hT(v) = hU(v). Suppose thenthere is a tiling X such that T ≤ X ≤ U. Then for all v, we musthave hT(v) ≤ hX(v) ≤ hU(v). Then for v 6= u, we have hT(v) =

hX(v) = hU(v), and for u we have hT(v) ≤ hX(v) ≤ hT(v) + 4, or0 ≤ hX(u)− hT(v) ≤ 4. Since hX(u)− hT(v) = 4k for some integerk, we must have hX(u)− hT(u) = 0 or 4, or hX(u) = hT(u)orhU(u).Therefor, for all v, we have hX(v) = hT(v) or hU(v), and so by Theo-rem 74, X = T or X = U.

Theorem 79 (Rémila (2004), First part of Proposition 7). Suppose Tand U are tilings of a region, and that T < U. Then we can perform an upflip on T to obtain a new tiling X such that T < X ≤ U.


dominoes ii 75

Proof. Take the smallest vertex v such that hT(v) < hU(v). (Thisimplies hT(v) ≤ hU(v)− 4.)

Now let v′ be a neighbor of v such that sp v′ = −1. If a dominodoes not cross the edge v− v′, then by Theorem 74, hT(v′) = hT(v)−1 < hU(v)− 1 ≤ hU(v′) + 1− 1 = hU(v′). So hT(v′) < hU(v). But thenv cannot be the smallest vertex, and so a domino must cross v− v′.

This is true for both neighbors of v with sp v′ = −1, and so wehave a flippable pair. Moreover, it is anticlockwise, and so we canperform an upflip to obtain tiling X. By theorem 76 we have hX(v) =hT(v) + 4 ≤ hT(v), and for all other vertices u we have hT(v) =

hX(u) ≤ hU(u). Thus, we have T < X ≤ U.

Let us write T < U when T ≤ U and T 6= U. A covering relation≺ is a such that if T ≺ U, then T < U, and there is no element Xsuch that T < X < U (Davey and Priestley, 2002, 1.14, p.11).

Theorem L1. (Davey and Priestley, 2002, 1.14, p.11) If T ≤ U are tilingsof a simply connected region, then one of the following hold:

(1) T = U.

(2) T ≺ U.

(3) There are tilings Vi such that T ≺ V1 ≺ V2 ≺ · · · ≺ U.


Proof. Suppose that T < U. Then either there is a tiling X such thatT < X < U, or there is not. In the latter case, T ≺ U. In the formercase, repeat the argument with T and X, and with X and U, untilwe recover the sequence T ≺ V1 ≺ V2 ≺ · · · ≺ U. The processmust end because no tiling is repeated (at each stage, we can writeT < X1 < · · · < U for the tilings found so far) and there are only afinite number or tilings.

Theorem 80. If R is a simply-connected region with tilings T and U, thefollowing are equivalent:

(1) T ≺ U

(2) T and U differ by a flippable pair, and the pair is anticlockwise in T.


Proof. Suppose T ≺ U, which means T < U. Then by Theorem 79

there is a tiling X that differs by a flip from T such that T < X ≤ U.If X 6= U, then we have T < X < U, which contradicts T ≺ U, and soX = U, which means U differs from T by a single flip.

76 polyominoes

(a) (b) (c) (d) (e)

Figure 78: Examples of Hasse dia-grams for the tilings of various regionswithout holes.

Suppose T and U differ by a flippable pair, anticlockwise in T.Now suppose X is a tiling such that T ≤ X ≤ U. Then by Theorem 78

either X = T or X = U, and so T ≺ U.

Alternative proof of second part. Suppose that for some V, we haveT ≤ V ≤ U. This means, that T ≺ V1 ≺ · · · ≺ Vk ≺ U (by TheoremL1), where one of Vi equals V, or, if we let V0 = T and Vk+1 = U, wehave V0 ≺ V1 ≺ · · · ≺ Vk ≺ Vk+1. Let fi be the flippable pair thattakes Vi−1 to Vi, and g the flippable pair that takes T to U.

Now consider the top-most (and left-most if there is more thanone) flippable. The top left corner is not moved by any other flip-pable pair. This corner must then be different in both T and U, andso must be moved by g. And in fact, this flippable pair must equal g.Similarly, we can show bottom-most, left-most and right-most flipsfrom fi must all equal g. But then all fi must equal g, but fi 6= fi+1

(the second one is already clockwise). So this situation is only possi-ble if there are no Vi. Thus, no V can lie between T and U.

Theorem 81 (Rémila (2004), Proposition 7). Suppose T < U. Then thereis a sequence of up flips that takes T to U.

[Not referenced]

Proof. This follows from combining Theorem L1 with Theorem 80, orby induction from Theorem 79.

dominoes ii 77

Figure 79: The flip graph of tilings ofthe 4× 4 square.

78 polyominoes

Theorem 82. If A ≺ T, B ≺ T, A ≺ U and B ≺ U, then either A = B orT = U.

[Not referenced]

Proof. Let Af1−→ T

f2−→ B and Ag1−→ U

g2−→ B. If f1 and f2 are the same,then A = B. If they are different, then f1 moves at least 2 cells that isnot moved by f2. Those same cells must be moved by g1 (they cannotbe moved by g2 because of the orientation). But then f1 = g1, andf2 = g2, and so T = U.

This theorem is generalized in Theorems L2 and L4.

Theorem 83. Suppose that A ≤ T, A ≤ U, B ≤ T, B ≤ U, and A 6= B.Then there exist a tiling V such that A ≤ V, B ≤ V, V ≤ T and V ≤ U.


Proof. If T ≤ U, we can choose V = T. If U ≤ V we can chooseV = U. Suppose then T and U cannot be compared.

Now A and B differ by closed strips, and say the outer ones are Si

and they surround vertices vj. Now consider the tiling X which hasall these strips in clockwise orientation. Clearly, A ≤ X and B ≤ X.But clearly X ≤ T and X ≤ U.

Theorem L2. Suppose Q is the set of all tilings bigger than both A and B.Then there exist a unique tiling V in Q smaller or equal to any element inQ.


Proof. Let Q = {X1, X2, · · · , Xn}. Now let V1 be a tiling bigger thanA and B but smaller than X1 and X2 (such a tiling must exist byTheorem 83). Let Vi be a tiling bigger than A and B but smaller thanVi−1 and Xi+1 (again, such a tiling exists by Theorem 83). Then Vn−1

is in Q, and Vn−1 ≤ Xi for all i.To prove uniqueness, suppose there is another element V′ such

that it is bigger than A and B and smaller than all elements in Q.In particular, V′ ≤ V. But V is smaller than all elements of Q, andV′ ∈ Q, so V ≤ V′. Therefor, V′ = V (by Theorem 77.2).

We call V the join of A and B and denote it by A ∨ B. The follow-ing is basically a restatement of the definition in a more convenientform.

Theorem L3.

(1) T ≤ T ∨U

dominoes ii 79

(2) U ≤ T ∨U

(3) If T ≤ A and U ≤ A, then T ∨U ≤ A.


Theorem L4. Suppose Q is the set of all tilings smaller than both A and B.Then there exist a unique tiling V in Q bigger or equal to any element in Q.


Proof. The proof is almost exactly like the one in Theorem L2.

We call V the meet of A and B and denote it by A ∧ B. As before,the following is basically a restatement of the definition in moreconvenient form.

Theorem L5.

(1) T ∧U ≤ T

(2) T ∧U ≤ U

(3) If A ≤ T and A ≤ U, then A ≤ T ∧U.


A set with a partial order for which the join and meet alwaysexists is called a lattice.

Theorem L6. Let T, U, and V be tilings of a region. The join and meetoperations satisfy the following laws (Davey and Priestley, 2002, Thereom2.9, p. 39):

T ∨U = U ∨ T T ∧U = U ∧ T Commutative (1)

(T ∨U) ∨V = T ∨ (U ∨V) (T ∧U) ∧V = T ∧ (U ∧V) Associative (2)

T ∨ (T ∧U) = T T ∧ (T ∨U) = T Absorption (3)

T ∨ T = T T ∧ T = T Idempotent (4)

[Not referenced]

Proof. Here we prove only the first of each of the four pairs of equa-tions. The remainder is proven with dual arguments using (TheoremL5).

(1) This follows directly from the symmetry in the definition of ∨.

80 polyominoes

(2) Let A = (T ∨ U) ∨ V. Then, (T ∨ U) ≤ A (Theorem L3.1),and V ≤ A (Theorem L3.2). From the first of these, we haveT ≤ A and U ≤ A. But then U ∨ V ≤ A (Theorem L3.3),and so T ∨ (U ∨ V) ≤ A (Theorem L3.3), or T ∨ (U ∨ V) ≤(T ∨U) ∨ V. Similarly, (by putting B = T ∨ (U ∨ V)) we canshow (T ∨ U) ∨ V ≤ T ∨ (U ∨ V). Taken together, we haveT ∨ (U ∨V) = (T ∨U) ∨V (Theorem 77.2).

(3) T ∧U ≤ T and T ≤ T, so T ∨ (T ∧U) ≤ T. But T ≤ T ∨ (T ∧U), so T ∨ (T ∧U) = T (Theorem 77.2).

(4) T ≤ T ∨ T (Theorem L3.1), and (since T ≤ T) T ∨ T ≤ T(Theorem L3.3). Thus, T = T ∨ T (Theorem 77.2).

Theorem 84. 4 Let T and U be tilings of a region. Then 4 This is essentially Rémila (2004,Proposition 5)

(1) min(hT(v), hU(v)) = hT∧U(v)

(2) max(hT(v), hU(v)) = hT∨U(v)


Proof. We will only prove (1), (2) can be proven symmetrically.First, let us show that f (v) = min(hT(v), hU(v)) corresponds to a

tiling. We prove the three conditions of Theorem 75.

(1) f (v∗) = min(hT(v∗), hU(v∗)) = min(0, 0) = 0.

(2) Consider neighbors u and v such that sp(u, v) = 1, and as-sume hT(u) < hU(u), or hT(u) ≤ hU(u) − 4. We also havehT(v) ≤ hT(u) + 1, and hU(v) ≥ hU(u) − 3, so hT(v) ≤hT(u) + 1 ≤ hU(u)− 4 + 1 ≤ hU(v) + 3− 4 + 1 = hU(v).

This proves if hT(u) < hU(u), then f (v) = hT(v). But, sincef (u) = hT(u), we have f (v)− f (u) = hT(v)− hT(u), whichmeans the second condition is satisfied. Similar argumentshold when hT(u) > hU(u) or hT(u) = hU(u).

(3) The above also shows the third condition is met when u and vlie on the border.

Therefor, f corresponds to a tiling.Now let this tiling be X. Since f = hX = min(hT , hU), we have

X ≤ T, U. Suppose then there is a tiling Y such that X ≤ Y ≤T, U. Then we have min(hT , hU) = hx ≤ hy ≤ hT , hU , which ispossible only if hY = hX , and therefor Y = X. Therefor X is themaximum tiling smaller than T and U, and therefor X = T ∧ U.Thus, min(hT(v), hU(v)) = hT∧U(v).

dominoes ii 81

Theorem L7. Let R be a region with a tiling. Then

(1) There exist a unique tiling E such that T ≤ E for all tilings T of R.

(2) There exist a unique tiling Z such that Z ≤ T for all tilings T of R.

(3) E = Z if only if R has a unique tiling.


Proof. Define E =∧

Ti. Clearly, Ti ≤ Z for all i. Suppose anothertiling E′ has this property. We have both E ≤ E′ and E′ ≤ E, so E = E(Theorem 77.2).

Similarly, define Z =∨

Ti. Z ≤ Ti for all i. And if another tiling Z′

has this property, we have both Z ≤ Z′ and Z′ ≤ Z, and so Z = Z′

(Theorem 77.2).Suppose R has a unique tiling T. Then clearly T = Z = E. Suppose

on the other hand E = Z, and consider any tiling T of R. Then wehave Z ≤ T ≤ E, or Z ≤ T ≤ Z, which implies T = Z (Theorem 77.2),so all tilings of R are equal, and so R has a unique tiling.

The tiling E in the theorem above is called the maximum tiling,and is denoted by 1R (this assumes the tileset T is understood), or ifthe region is understood, simply by 1. Similarly, the tiling Z is calledthe minimum tiling, and it is denoted by 0R, or simply 0, if R isunderstood. The minimum and maximum tilings for some rectanglesare shown in Figure 80.

Figure 80: The minimum and maximumtilings for small rectangles.A distributive lattice is a lattice that satisfies the following two

distribution properties (Davey and Priestley, 2002, Def. 4.4, p. 86):

T ∨ (U ∧V) = (T ∨U) ∧ (T ∨V) (4.1)

T ∧ (U ∨V) = (T ∧U) ∨ (T ∧V) (4.2)

82 polyominoes

Theorem 85 (Thurston (1990)). The set of tilings of a simply-connectedregion forms a distributive lattice.

[Not referenced]

Proof. We need to prove that any three tilings T, U, and V satisfy thedistributive properties in Equations 4.1 and 4.2.

For the first equation, we use Theorem 84 to find an expression forthe tilings on the left hand and right hand of the expression.

Let X = T ∨ (U ∧ V). This gives us hx = max(hT , min(hU , hV)) =

min(max(hT , hU), min(hT , hV)).Let Y = (T∨U)∧ (T∨V). Then hY = min(max(hT , hU), min(hT , hV)),

so hX = hY, and so X = Y by Theorem 74.A similar argument can be used to show the second equation is

also satisfied.

Problem 29. Find an algorithm to find the minimum and maximum tilingsof a region given an arbitrary tiling of the region.

Theorem 86. Suppose that T ≺ X1 ≺ · · · ≺ Xm ≺ U and T ≺ Y1 ≺· · · ≺ Yn ≺ U. Then m = n.


Proof. Let ui be all the vertices of R. Let vi be the vertex that changesheight when we go from Xi−1 to Xi (and let X0 = T and Xm+1 = U).Similarly, let wi be the vertices as we move along Yi. Then we have∑i hU(ui)− hT(ui) = ∑i=1 m + 1vi = ∑i=1 n + 1wi, or ∑i=1 m + 14 =

∑i=1 n + 14, or 4(m + 1) = 4(n + 1), which means m = n.

The rank ρ(T) of a tiling T is defined as follows5: 5 Rank functions can be defined forgeneral partially orders, see for exampleStanley (1986, p. 99).(1) ρ(0) = 0.

(2) If T ≺ U, then ρ(T) + 1 = ρ(U).

Theorem L7 ensures that 0 exists, and Theorem 86 ensures this defini-tion is consistent: it is not possible to arrive at different rank depend-ing on the chain chosen.

Note that if T ≤ U, then ρ(T) ≤ ρ(U).

Theorem 87. Suppose that A, B ≺ X, and A 6= B. Then there is a tiling Ysuch that Y ≺ A, B.


dominoes ii 83

Proof. Because A, B ≺ X, and A 6= B, A and B differ by two flips,and that these flips do not overlap (if they did, then A ≺ X ≺ B, orB ≺ X ≺ A, which is not the case). Suppose we perform flip f fromA to X, and g from X to A. We can then perform g on A to obtain Y,and f on Y to obtain B, which shows that Y ≺ A, B.

Theorem 88. Let µ(T, U) be the minimum number of flips necessary toturn T into U. Then µ(T, U) = 2ρ(T ∧U) − ρ(T) − ρ(U) = ρ(T) +ρ(U)− 2ρ(T ∨U).


Proof. Suppose a sequence of tilings from T to V via flips is U1, U2, · · ·Uk.A triangle on A and B is a sequence of tilings such thatA ≺ X1 ≺ X2 ≺ · · · ≺ Xm ≺ C, and B ≺ Y1 ≺ Y2 ≺ · · · ≺ Ym ≺ C.Now there is a tiling Ui such that Ui ≺ Ui−1, Ui+1 (unless this

sequence is a triangle on T and U). By Theorem 87 there is a U′i suchthat Ui−1, Ui+1 ≺ Ui. We can repeat this procedure until we have atriangle. Note that the operation does not change the number of flips.

Now let C be the top of this triangle. Then T, U ≤ C, thus T ∨U ≤C. And since T, U ≤ T ∨U, by Theorem we have a flip path from T toU via T ∨U, and this path of length 2ρ(T ∧U)− ρ(T)− ρ(U). Andsince T ∨U ≤ C, this path must be smaller than the path through C,and hence, the shortest path via flips between two tilings is throughthe join, given by 2ρ(T ∧U)− ρ(T)− ρ(U).

A symmetric argument shows another shortest path is through themeet, which gives us the length ρ(T) + ρ(U)− 2ρ(T ∨U).

So µ(T, U) = 2ρ(T ∧U)− ρ(T)− ρ(U) = ρ(T) + ρ(U)− 2ρ(T ∨U).

Theorem L8 (Parlier and Zappa (2017)). Let T and U be two tilings of aregion. Then µ(T, U) ≤ ρ(1).

[Not referenced]

Proof. Suppose that ρ(T) + R(U) > ρ(1). By Theorem 88 µ(T, U) =

2ρ(1)− (ρ(T) + ρ(U)), so µ(T, U) < ρ(1). Suppose on the other handthat ρ(T) + ρ(U) ≤ ρ(1). By Theorem 88, µ(T, U) = ρ(T) + ρ(U)−2ρ(T ∨U) ≤ ρ(T) + ρ(U) ≤ ρ(1).

We next give the maximum ranking for rectangles and Aztec dia-monds. Using the above theorem, we know what is the most flips weneed to turn one tiling of one of these figures into another tiling.

84 polyominoes

Theorem 89 (Parlier and Zappa (2017)). For rectangles,

ρ(1R(m,n)) =

mn2

4 −n3

12 −n6 for n even

mn2

4 −n3

12 + n12 −

m4 otherwise

. (4.3)

For squares,

ρ(1R(n,n)) =n3 − n

6. (4.4)

[Not referenced]

Table 7 shows some values for ρ(1R(n,n)).

2 4 6 8 10 12

1 0 0 0 0 0 0

2 1 3 5 7 9 11

3 2 6 10 14 18 22

4 3 10 18 26 34 42

5 4 14 26 38 50 62

6 5 18 35 53 71 89

7 6 22 44 68 92 116

8 7 26 53 84 116 148

9 8 30 62 100 140 180

10 9 34 71 116 165 215

11 10 38 80 132 190 250

12 11 42 89 148 215 286

Table 7: ρ(1) for R(m, n).

Theorem 90 (Parlier and Zappa (2017)). For Aztec diamonds,

ρ(1A(n)) =n3

3+

n2

2+

n6

[Not referenced]

Table 8 shows some values for ρ(1A(n)).n A000330

1 1

2 5

3 14

4 30

5 55

6 91

7 140

8 204

9 285

10 385

11 506

12 650

Table 8: ρ(1) for A(n).

The following relates the height function of a tiling with its rank-ing.

Theorem 91. The ranking of a tiling is given by

ρ(T) =14

(∑

ihT(vi)−∑

ih0(vi)

).

[Not referenced]

Proof. Clearly, ρ(0) = 0. Suppose now the above is true for all tilingsT with ρ(T) < k. Suppose now T is any tiling with rank ρ(U) =

ρ(T) + 1. This means, that U and T differ by a flip, and so there is avertex vm such such hU(vm) = hT(vm) + 4, and hU(v) = hT(v) for allother vertices. Thus

ρ(U) = ρ(T) + 1

=14

(∑

ihT(vi)−∑

ih0(vi)

)+ 1

=14

(−4 + ∑

ihU(vi)−∑

ih0(vi)

)+ 1

=14

(∑

ihU(vi)−∑

ih0(vi)

).


dominoes ii 85

A local minimum is a vertex v with neighbors v′, such that h(v) <h(v′) for all neighbors v′. Similarly, a local maximum is a vertex vsuch that h(v) > h(v′) for all its neighbors v′.

Theorem 92 (Thurston (1990)). Algorithm for finding a tiling of a regionif it exists.


Proof.

(1) An interior vertex is a local minimum or maximum if and onlyif it lies at the center of a flippable pair.

(2) The minimal tiling of a figure has no local maximums in theinterior.

(3) If it did, we could get a tiling smaller than the minimum byperforming a flip.

(4) Therefor, the vertices for which h0(v) is maximum lie onthe border. Let v be such a vertex, and let u and w be itsneighbors so that u, v and w is in clockwise order. Note thath0(u), h0(w) < h(v), so the edge between u and v must bewhite, and the edge between v and w must be black.

(5) Therefor v cannot be a corner, which means u, v and w lie in astraight line.

(6) There cannot be a domino path from v to the interior neigh-boring vertex v′, otherwise it will be higher than the maximum(since it has a black sell on the left). Therefor, a domino mustcover the edge between v and v′. We found one domino of thetiling 0 (if it exists).

(7) We can now remove the two covered cells, update the heightfunction on their edges, and repeat the process. If, whenwe update the height function we cannot assign consistentheights, a tiling is impossible.

A direct path from one vertex to another is a path from the onevertex to the other so that it always has a black cell on the left. Notethat this path is not symmetric (See Figure 81). The distance betweentwo vertices is the length of the shortest direct path between them(note that since the path has black cells on the left, the spin from onevertex on the path to the next is always 1).

Figure 81: A direct path between twonodes. If we change the direction of thepath, some edges are different.

86 polyominoes

Removing a cylinder from a region does not affect the heights ofthe vertices of the border that remain. To make this statement precise,we need a function that maps the cells of the reduced region to thecells of the original region. If R is a region, and S is a 2-cylinder, thenfS : R S → R maps the cells in R S to their original counterpartsin S.

Theorem 93. If R is a region and S is a removable 2-cylinder, and h aheight function. Then hRS(vi) = hR( fS(vi)) for all vertices.

[Not referenced]

Proof. Let vk be a vertex on the border of R− S, with height functiondefined relative to the vertex v0. Let v0, v1, v2, · · · vk be a path thatdoes not cross any dominoes. The cylinder can cut through this pathzero times, once, or twice.

(1) If it does not cut through the path, then the height of vk is thesame in both R and R S.

(2) If it cuts through the path once, there is one extra white andone extra black edge in R, which means the net effect is 0.

(3) If it cuts through the path twice, there is two extra white andtwo extra black edges, and the net effect on the height is 0.

In all cases, hRS(vk) = hR( fS(vk))

Theorem 94. Let S be a removable cylinder of R, and suppose u and v arevertices with not in the interior of the cylinder, or the borders of the cylindershared with R. Then

dRS(u, v) < dR( f (u), f (v)) < dRS(u, v) + 4.

[Not referenced]

Proof. If u and v are on the same side of the cylinder, the shortestpath between them is unaffected, and therefor the length is the same.

If they lie on opposite sides of the cylinder, we can find a shortestpath between them that has at most 4 edges in the cylinder. Thisis because a piece of the path with more than 4 edges must have asection of 3 consecutive edges where the path goes around a cell, andcan be changed to go around a different cell outside the cylinder. SeeFigure 82 for an example. And when we remove the cylinder, thispath is at most 4 units shorter.

Figure 82: Rerouting a direct path.Problem 30.

dominoes ii 87

(1) The two theorems above show us that if |hRS( f (x), f (y))| ≤dRS( f (x), f (y)), then |hR(x, y)| ≤ dR(x, y). Use this to show thatif R S is tileable, then so is R. You also need to take the vertices ofS into account. This is a different way to prove Theorem 37.

(2) Show that the converse does not hold by checking that Figure 39provides a counter example.

Theorem 95. A simply-connected region R is tileable by dominoes if andonly if ∆(R) = 0 and |h(x, y)| ≤ d(x, y) for all vertices x and y on theborder.


Proof. Suppose that there are nodes x and y such that h > d. Let S bea subregion of R between a shortest direct path between x and y andthe border of R.

If R is tileable, then φ(S) = ∆(S) = h+d4 . But the only place where

dominoes can cross is on the direct path between x and y. Note thatbecause of the paths shape, we have

d3≤ φ(S) ≤ d

2. (4.5)

Thus h+d4 ≤ d

2 , or h + d ≤ 2d, or h ≤ d, which contradicts ourassumption, and therefor R is not tileable.

On the other hand, suppose R is not tileable. Then there is a badpatch S. This bad patch cannot lie completely in the interior of theregion (because all such cells have 4 neighbors in R, and thereforcannot be a bad patch). So suppose the bad patch shares a borderwith R between vertices x and y. WLG assume the patch is black(if it is white, we can also find a black bad patch by Theorem 33).Now take a shortest direct path between x and y, and let S′ be thesubregion between this path and the border of R, with φ(S′) > ∆(S′)(This is possible since the region is not tileable, and either the fluxmust be bigger or smaller; since there are two possible subregion, wecan find one with the necessary constraint.)

By Equation 4.5 we have h+d4 < φ(S′) ≤ h

2 , so h + d < 2h, ord < h.

The section Further Reading gives some references in dealing withregions with holes, which we don’t cover here.

4.1.2 Forced flippable pairs

The following section is from Kranakis (1996). A pyramid6 is stack 6 This is a slightly different definitionthan given in Kranakis (1996), whichdoes not account correctly for all thecases in the pyramid lemma.

polyomino with the following forms:

88 polyominoes

• For an even number of 2k columns: B(3 · 4 · 5, · · · (k + 2)2 · · · 5 ·4 · 3). See Figure 83.

• For an odd number of 2k + 1 columns: B(3 · 4 · 5, · · · (k +

2)3 · · · 5 · 4 · 3). See Figure 84.

Figure 83: An even pyramid with vectorB(3 · 4 · 5 · 6 · 72 · 6 · 5 · 4 · 3).

Figure 84: An odd pyramid with vectorB(3 · 4 · 5 · 63 · 5 · 4 · 3).

Figure 85: An L-configuration.

Figure 86: An R-configuration.

An L-configuation is two dominoes at 90 degrees that form a“left corner”, (Figure 85); an R-configuration is two dominoes at 90

degrees that form a “right corner” (Figure 86).If each of these occur on the same height within a tiling with

the L-configuration on the left, we call the combination an LR-configuration.

Figure 87: This tiling has several LR-configurations. One is shown in yellow,another one in red.

Theorem 96 (The Pyramid Lemma, Kranakis (1996), Lemma 4). Ifa domino tiling of a rectangle is given, and a LR-domino configuration is

dominoes ii 89

present in the tiling, then there is a flippable pair present in the smallestpyramid that contains the LR-configuration on its base.


Proof. There are a few cases to consider, but the general pattern is thesame. Take the LR configuration shown. The left horizontal dominocan either be covered by another horizontal domino (in which casewe have a flippable pair), or by a vertical domino. This verticaldomino either has another vertical domino to the right (in whichcase we have a flippable pair) or a horizontal domino (forming an L-configuration). A similar argument on the right shows we either havea flippable pair, or another R-configuration. We continue this process,until eventually a flippable pair is forced.

Figures 88 and 89 shows all the different cases for different LR-configurations and whether the smallest pyramid is even or odd.

Figure 88: Forced flippable pair in evenpyramids.

Figure 89: Forced flippable pair in oddpyramids.

Theorem 97. Suppose there is a vertical domino on the bottom border ora rectangular tiling, with no other vertical dominoes between it and the left(or right) border of the rectangle. Then there is a flippable pair inside thepyramid with the vertical domino in one corner, and the other corner justoutside the left (or right) rectangle border.


90 polyominoes

Proof. The logic is exactly the same as used in the proof of Theorem96: If we try to fill the tiling without adding any flippable pairs, weare eventually forced to add a flippable pair. Figure 90 shows thesetup.

Figure 90: A forced flippable pair in thepyramid between a vertical domino andthe rectangle border.

Theorem 98. If there is an R-configuration in a tiling if a rectangle, andwe draw a diagonal to the top left, there is a flippable pair that crosses thisdiagonal.


Proof. Again, the logic is the same as in Theorem 96. If we fill in thetiling by avoiding flippable pairs, we are eventually forced to add aflippable pair. Figure 91 shows the setup.

Figure 91: Forced flippable pair occurson the diagonal that starts from theR-configuration.

Theorem 99 (Kranakis (1996), Theorem 5). Domino tilings of rectangleswith both sides of length greater than 1 must always have a flippable pair.


Proof. This follows directly from Theorem 65, since if the rectanglehas all sides greater than 1, it has no peaks, and also it has no holes.

We give here an alternative proof given in Kranakis (1996).Suppose the region rectangle is R(m, n). We prove it by induction

on the area mn.If m = 2, then either the first two cells are covered by two horizon-

tal dominoes (a flippable pair), or by a vertical domino as in Figure101. In the latter case, the next two cells are either covered by twohorizontal dominoes (forming a flippable pair), or a vertical domino,forming a flippable pair with the first vertical domino. In all cases wehave a flippable pair, and by symmetry the same is true if n = 2.

dominoes ii 91

Suppose then both m, n > 2. We know (Theorem 1) that mn iseven, so either m or n must be even. WLG suppose m (the horizontaldimension) is even. If there are now vertical dominoes touchinga horizontal border, we can strip off a row to get a new rectangleR(m, n − 1) with area m(n − 1) < mn, and the theorem follows byinduction.

Suppose there are vertical dominoes along the (say bottom) border.Since m is even, there must be at least two (Theorem 21). Choose twoadjacent such dominoes. Then all the dominoes next to the borderare horizontal, and by Theorem 96 there is a flippable pair in thepyramid induced by the corner configurations.

Problem 31. Show that if a tiling of R(m, n) with either m > 3 or n > 3has two overlapping flippable pairs in a subregion S, then it has two flip-pable pairs that don’t overlap.

Theorem 100.

(1) In a rectangle, there are at most two tilings with exactly only oneflippable pair.

(2) If there are such tilings, the one is a rotation of the other.

(3) R(m, n) have 0 such tilings if m > n + 2 or n > m + 2.

(4) R(m, n) has 1 such tiling if m = n + 2 or n = m + 2.

(5) R(m, n) has 2 such tilings otherwise, that is m < n + 2 and n <

m + 2, or n− 2 < m < n + 2.


Proof. A tiling with only one flippable pair must be either minimal ormaximal (since its rank can only be reduced or increased by flippingthe flippable pair, it cannot be both increased and decreased sincethere is only one flippable pair.) Since there is exactly one minimaland maximal tiling (Theorem L7), there can be at most two tilingswith flippable pairs.

Suppose there are two tilings with exactly one flippable pair.

(1) If m = n, we have a square. Now if we rotate one tiling by 90°,we get a different tiling with one flippable pair. But there canbe at most two, so this tiling must equal the other tiling, andso the two tilings are rotations of each other.

(2) If m 6= n, suppose one tiling is not symmetric w.r.t. a 180°turn. Then by turning it, we can get another tiling. Again,there must be at least two, so it must equal the other tiling.

92 polyominoes

If both are symmetric with respect to a 180° turn, then theymust be equal (?), but then there cannot be two tilings withone flippable pair.

So in all cases, there if there are two tilings with one flippable pair,the two tilings are rotational copies of each other.

R(2, m) has more than two flippable pairs for m > 4. Each verticaldomino must be adjacent to another vertical domino, forming a flip-pable pair. But if there is only one such, all other dominoes must behorizontal, and so if m > 4, there must be at least 3, and they and anyarrangement must result in at least one more flippable pair.

R(3, m) must have a fault by Theorem 118. If the fault is vertical,we have R(2, m) subtiling, which must have at least two flippablepairs (as we proved above). If the fault is horizontal, we have tworectangles with height at least two, and so neither has a peak and soeach must have a flippable pair (Theorem 65), giving the total figureat least two flippable pairs.

Theorem 101 (Kranakis (1996), Corollaries 7 and 8). For even squaresR(n, n), there are

(1) A unique tiling (up to rotation) that has 1 flippable pair.

(2) A unique tiling (up to rotation) that has 2 flippable pairs.

[Not referenced]

Proof. We showed that any rectangle has at most 2 tilings with 1

flippable pair, and they are rotations of each other (Theorem 100).(This first part is a different proof from the one given in Kranakis(1996)).

For the second part, we consider several cases.

(1) Suppose there are 4 vertical dominoes on the border. Some ofthese may be adjacent to form groups.

(a) If there are 4 groups, there are 3 flippable pairs by Theorem96.

(b) If there are 3 groups, one group has a flippable pair, andthere are 2 additional flippable pairs by 96.

(c) If there are 2 groups:

i. If one group is not in the corner, we have one flippablepair in one of the groups, one flippable pair betweenthem (Theorem 96), and one flippable pair between theone group and rectangle border (97).

dominoes ii 93

ii. If both groups are in the corner, and both are flippablepairs, we have a third flippable pair between them (Theo-rem 96). If they are not both flippable pairs, then one is aflippable pair, and two more flippables are forced on thediagonals shown in Figure 92.

Figure 92: Two additional flippablepairs are forced.

In all cases, if we have 4 vertical dominoes, there are at least 3

flippable pairs, so this situation cannot occur.

(2) Suppose then there are 2 vertical dominoes.

(a) If there is one group, we have a flippabe pair. If it is notin the corner, we have two additional flippable pairs byTheorem 97, if it is, we have two additional flippable pairson the diagonals as shown in Figure 93.

Figure 93: Two additional flippablepairs are forced.

(b) If there are two groups

i. If neither is in the corner, we have a flippable pair be-tween them, and between each and a rectangle border,giving three in total.

ii. If one domino is in the corner, we have have one flip-pable pair between them, and one between the other andthe rectangle border. There is a third flippable pair forcedby the top corners of the rectangle, and this flippable paircannot coincide with either of the other two. See Figure94.

Figure 94: 3 forced flippable pairs.

This only leaves the case where there is a vertical domino ineach corner.

The final case if there are no vertical dominoes.

We have shown that if there are exactly two flippable pairs, we ei-ther need no vertical dominoes, or there are exactly two, one in eachcorner. This can only be realized in one of two ways.

Figure 95: Two ways to complete theborder.

In either case, the two flippable pairs are in a sub-square R(n −2, n − 2), and we can repeat the process above on this sub-square.

94 polyominoes

Going forward, we only have one option to build the border eachtime.

(a) A legal way to complete theborder in the next step.

(b) Not a legal way to complete theborder in the next step.

Figure 96: The next step.

Eventually, we arrive at one of the two tilings of the 4× 4 squareshown in Figure 97. This shows there are only two tilings with 2

flippable pairs, and they are 90 degree rotations of each other.

Figure 97: Two ways in which a 4× 4square can have exactly two flippablepairs.In Kranakis (1996, Theorem 9) the following claim is made: In a

tiling of a n × n square with exactly 3 non-overlapping squares, there is a8× 8 square in the center that contains the three non-overlapping squares.Although a proof is given the claim is in fact false, as the counterex-ample in Figure 98 shows. This pattern can be also constructed forlarger squares, showing there is no bound on the subregion that con-tains the three squares.

The author also conjectures that in tilings with exactly four squares,either all the tilings lie within a bounded square, or the tiling can bepartitioned into four squares with has a flippable pair in the center ofeach. This is also false, as shown in Figure 99.

It does seem plausible that in a tiling with more than one flippablepair, there is not a square larger than R(n/2, n/2) that contains onlyone flippable pair.

Theorem 102 (Kranakis (1996), Theorem 10). A domino tiling of arectangle R(m, n), with 1 < m ≤ n must have at least

⌊ nm+1

⌋non-

overlapping flippable pairs.

[Not referenced]

Proof. The argument is similar to that used for Theorem 99, usingTheorems 96 and 97.

Figure 100 shows several tilings that achieves the minimum.

dominoes ii 95

Figure 98: A counterexample toKranakis’s Theorem.

Figure 99: A counter example toKranakis’s Conjecture.

96 polyominoes

Figure 100: Tilings that achieves theminimum number of flips.

4.2 Counting Tilings

We have already seen a few results about the number of tilings ofcertain regions:

• Snakes have a unique tiling.

• Rings have two tilings.

• Any region with a 2× 2 square as subregion has at least twotilings.

In this section we look at the number of tilings of rectangles andAztec diamonds. We will denote the number of tilings of a region Rby a tileset T as #T R, or if T is understood, simply #R.

4.2.1 Rectangles

We will use the notation R(m, n) for an m× n rectangular region.

Theorem 103. Let Fn be the Fibbonacci numbers 1, 1, 2, 3, 5, 8, · · · (A000045).

#R(2, n) = Fn+1 =1√5

(1 +√

52

)n+1

−(

1−√

52

)n+1 . (4.6)



dominoes ii 97

Proof. There are two ways in which to start a tiling: with a verticaldomino, or two horizontal dominoes.

The first case leaves us with a R(2, n − 1) and the second caseleaves us with a R(2, n− 2). From this we find the recursion: #R(2, n) =#R(2, n− 1) + #R(2, n− 2). And by checking that #R(2, 1) = 1 and#R(2, 2) = 2, we are lead to the familiar Fibonacci sequence. That is,#R(2, n) = Fn+1. See Knuth et al. (1989, Section 6.6) for how to derivethe expression for Fibonacci numbers.

Figure 101: The two ways a dominotiling of R(2, n) can start.

With a bit more work, we can also find a recursion for R(3, n).First, n must be even. That means we can divide the rectangle into

R(2, 3) blocks, and the only possibilities of how they are covered withdominoes are the ones shown in Figure 102.

(a) Univer-sal

(b) BlueTower

(c) RedTower

(d) BlueStart

(e) BlueMiddle

(f) BlueEnd

(g) RedStart

(h) RedMiddle

(i) RedEnd

Figure 102: The only blocks that canmake up R(3, n). The terminology isfrom Butler et al. (2010)

Theorem 104.

#R(3, n) =3 +√

33

(2 +√

3)n/2

+3−√

33

(2−√

3)n/2


Proof. Let Sn be the number of tilings possible for an 3× n rectanglestarting with either the blue or red start piece. We can only start arectangle with the universal piece, one of the towers, or one of thestart pieces. This gives us:

#R(3, n) = 3#R(n− 2) + 2Sn.

For a rectangle starting with a start piece, the only next piece iseither the corresponding middle piece, or the corresponding endpiece. This gives us:

Sn = #R(3, n− 4) + #R(3, n− 6) + . . . + #R(3, 2) + 1

Popping this into the equation above gives us:

98 polyominoes

#R(3, n) =3#R(3, n− 2) + 2(#R(3, n− 4) + #R(3, n− 6) + . . . + #R(3, 2) + 1)

#R(3, n)− #R(3, n− 2) =3#R(3, n− 2) + 2(#R(3, n− 4) + #R(3, n− 6) + . . . + #R(3, 2) + 1)

− 3#R(3, n− 4)− 2(#R(3, n− 6) + #R(3, n− 8) + . . . + #R(3, 2) + 1)

=3#R(3, n− 2)− #R(3, n− 4)

Which gives us:

#R(3, n) = 4#R(3, n− 4)− #R(3, n− 4) (4.7)

Solving7 this recurrence gives us: 7 See Further Reading for references onhow to do this.

#R(3, n) =3 +√

36

(2 +√

3)n/2

+3−√

36

(2−√

3)n/2

We get the sequence for even n: 3, 11, 41, 153, 571, . . . This isA001835 (skipping the first two terms).

The number of tilings of R(3, n) without the universal tile is givenby 2 · 3n/2−1 for even n (Butler et al., 2010).

Figure 103: All the ways R(4, n) canbegin.There are eight ways in which an R(4, n) tiling can begin, shown

and named in the picture below. We’ll count the number with eachkind of beginning configuration and add the results up to get #R(4, n).This gives us

#R(4, n) = #R(4, n− 1) + 5#R(4, n− 2) + #R(4, n− 3)− #R(4, n− 4),(4.8)

with #R(4, 0) = 1, #R(4, 1) = 1, #R(4, 2) = 5, and #R(4, 3) = 11.How this occurrence is derived is shown by Kass (2014). An ex-

plicit formula is given in the entry for this sequence: A005178.From 5 onwards it becomes very tedious to do this type of re-

currence derivation by hand. Figure 104 lists the 25 ways a 5 × nrectangle can begin.



dominoes ii 99

Figure 104: All the ways a n × 5-rectangle can begin.

100 polyominoes

Theorem 105 (Kasteleyn (1961)). The number of tilings of a m × nrectangle is given by

#R(m, n) =m

∏k=1

n

∏l=1

2

√cos2 kπ

m + 1+ cos2 lπ

n + 1(4.9)

[Not referenced]

The proof uses advanced ideas and is lengthy, so I omit it. For a read-able exposition of this theorem, see Stucky (2015). Table 9 tabulatessome values. Showing that this theorem is equivalent to Theorems103 and 104 for m = 2 and m = 3 is a tricky exercise.

A099390 1 2 3 4 5 6 7 8 9 10 11

1 0 1 0 1 0 1 0 1 0 1 0

2 1 2 3 5 8 13 21 34 55 89 144

3 0 3 0 11 0 41 0 153 0 571 0

4 1 5 11 36 95 281 781 2245 6336 18061 51205

5 0 8 0 95 0 1183 0 14824 0 185921 0

6 1 13 41 281 1183 6728 31529 167089 817991 4213133 21001799

7 0 21 0 781 0 31529 0 1292697 0 53175517 0

8 1 34 153 2245 14824 167089 1292697 12988816 108435745 1031151241

9 0 55 0 6336

10 1 89 571

11 0 144

Table 9: Number of domino tilings of am× n rectangle.

Problem 32. How many tilings does B(m2 · 2n) have? (See Figure 105.)

Figure 105: An example of a regionwith an L-shape composed from tworectangles of width 2.

4.2.2 Aztec Diamond

An Aztec diamond A(n) is a region defined as follows:

(1) A(1) = R(2, 2)

(2) To form A(n + 1), attach a neighboring cell to each open edgeof A(n).

It is easy to see that the area of an Aztec diamond is given by2n(n + 1) (if you divide the Aztec diamond into quarters, each quar-ter is a triangle with 1 + 2 + 3 + n · · · = n(n + 1)/2 cells).

Each row of an Aztec diamond is a strip polyomino with an evennumber of (connected) cells, and is therefor tileable by dominoes(Theorem 51). In fact, Aztec diamonds are strip polyominoes them-selves.


dominoes ii 101

Figure 106: Aztec diamonds A(1), A(2),A(3) and A(4).

Problem 33. Prove that Aztec diamonds are

(1) strip polyominoes

(2) Saturnian

The number of tilings of A(n) has been determined, but a proofof this is difficult, and I won’t give it here. Instead, I will prove someother interesting properties.

Theorem 106. Let R be an Aztec diamond, and S a subregion of R also anAztec diamond with radius one less, sharing some border with R. Then inany tiling of R, either 0 or 2 dominoes cross the border of S.

[Not referenced]

Proof. Since S and R− S is tileable, it is possible that a tiling of R hasno dominoes that cross the border of S. Suppose one cross the bor-der, and that the cell in S covered by that domino is white. Anotherdomino must cross with the cell in S covered by it black (to keep thebalance). So if one domino cross, then at least two must cross.

Any dominoes that crosses the border with the same color cellsin S, must lie on the same side of S, and therefor two with no otherborder-crossing dominoes between them demarcates a snake withendpoints of the same color, which then has odd area (Theorem 50)and is untileable (Theorem 1).

Figure 108 shows an example. Note also that most dominoes in thesubregion is aligned with similar orientation. This partially explainsthe Arctic behavior we discuss in the next subsection.

Let A(n, i, j) be an Aztec diamond with two cells on the borderremoved in columns i and j as in the theorem above. We can thenwrite:

#A(n + 1) = #A(n) + 2n

∑i=1

2n

∑j=n+1

#A(n, i, j). (4.10)

102 polyominoes

Figure 107: The 8 tilings of AD(2)

Figure 108: At most two dominoes cancross the border of the subregion.

Theorem 107. The number of domino tilings of an Aztec diamond A(n) isgiven by

#A(n) = 2(2n) = 2

n(n+1)2

[Not referenced]

The theorem was first proved in Elkies et al. (1992a) and Elkies et al.(1992b) (they give four different proofs). Simpler proofs are given inEu and Fu (2005) and Fendler and Grieser (2016), with more detailsprovided in Simon (2016). Table 10 tabulates some values.

n |A(n)| #A(n)A046092 A006125

0 0 1

1 4 2

2 12 8

3 24 64

4 40 1024

5 60 32768

6 84 2097152

7 112 268435456

8 144 68719476736

9 180 35184372088832

10 220 36028797018963968

11 264 73786976294838206464

12 312 302231454903657293676544

13 364 2475880078570760549798248448

14 420 40564819207303340847894502572032

Table 10: Number of tilings of the Aztecdiamond.



dominoes ii 103

4.2.3 Arctic Behavior

In Jockusch et al. (1998) the authors proved that randomly generatedtilings of large Aztec diamonds had some peculiar patterns in thecorners: the dominoes tend to be oriented in specific ways. Thisphenomenon does not occur for rectangles, where the orientation ofdominoes are arbitrary.

Figures 109 and 110 show examples.8 8 The software used to generate thissoftware is available online; see Doer-aene.

Why is this happening? The formal statement and proof of thistheorem (called the Arctic Circle Theorem) is beyond the scope of thisessay. But we can get of an intuitive understanding of what is goingon. We have already seen that in certain subregions next to the bor-der, dominoes tend up to line a certain way (Sections 4.1.2 and 3.2.2).Another way to look at this phenomenon is to look at the flow. Di-vide the region into four partitions as shown. In each, we calculatethe deficiency as 2(n+1)

4 . That means, we need as many dominoes tocross over the axes. The axis has 2

⌊ n2⌋+ 1 crossing points. A quick

calculation shows in this setup

c(n) =

2d(n) + 1 if n is even

2d(n)− 1 if n is odd.

This means about at least half the potential crossing points need tobe actually crossed. Now when two such dominoes lie next to eachother, the also determine the orientation of two dominoes betweenthem. If there are three next to each other, then they determine theorientation of 6; in general the more crossing dominoes lie next toeach other, the more other dominoes’ orientation are forced. Becausethere is a large amount of dominoes that need to cross relative tothe number of crossing points, on average there will be plenty ofgroups of adjacent dominoes oriented the same way, and therefor, onaverage, dominoes in this region tend to line up.

With this type of reasoning, we can see that this type of behaviorwill be typical of regions with reasonably sized stretches of border ofthe same color.

4.3 Constraints and Monominoes

4.3.1 Optimal Tilings

If a region is not tileable, how close can we get to tiling it? Askeddifferently: Given a set of dominoes and monominoes, what is theleast number of monominoes required to tile the region?

Let T be a tileset without any monominoes, and let T + be the tileset with a monomino added. Given some region R to tile with T +,

104 polyominoes

Figure 109: A tiling of an Aztec dia-mond.

dominoes ii 105

Figure 110: A tiling of an Aztec dia-mond.

106 polyominoes

the least number of monominoes we require is called the gap num-ber of the region, with respect to the tileset T (Hochberg, 2015), andwe will denote it by GT (R), or simply G(R) if the tile set is clear. (Inthis section, T is always the set containing the domino.) If G(R) = 0,then R is tileable.

A tiling of R that uses exactly G(R) monominoes is called optimal(following Bodini and Lumbroso (2009)).

Theorem 108. For dominoes, the gap number must at least be the absolutedeficiency:

G(R) ≥ |∆(R)|.


Proof. Let M be the set of cells that are covered by monominoesin a optimal tiling of R. Then R − M is tileable by dominoes, so∆(R − M) = 0 (Theorem 23), and |∆(M)| ≤ |M| = G(R) (SeeProblem 7.3). But |∆(R)| ≤ |∆(R − M)|+ |∆(M)| (Problem 7.4), so|∆(R)| ≤ G(M).

Theorem 109. For a tileset T where |P| = n for all P ∈ T , we have

GT (R) ≡ |R| (mod n). (4.11)


Proof. If GT (R) = g, then there is a tiling with tiles from T andg monominoes. Remove the cells covered by monominoes to formR′. Now since R′ is tileable by T , we know that |R′| ≡ 0 (mod n)(Theorem 1). Thus g + |R′| ≡ g (mod n), or |R| ≡ g (mod n).

Note that this theorem is stated for any tileset, not just for domi-noes. For dominoes, it follows that G(R) ≡ |R| (mod 2).

Theorem 110. If R has a tiling with dominoes and a single domino, thenG(R) = 1, and so the tiling is optimal.


Proof. G(R) ≤ 1 (since we have a tiling realized with one monomino),and it cannot be 0 since |R| must be odd (Theorem 109), and soG(R) = 1.

Theorem 111. Strip polyominoes with odd area has gap number 1.

[Not referenced]

dominoes ii 107

Proof. Put the monomino on the first cell of the strip polyomino. Theremainder of the figure is a strip polyomino with an even numberof cells, and therefor tileable by dominoes (Theorem 51). We havea tiling with single monomino, and so the gap number must be 1

(Theorem 110).

Theorem 112. Monominoes in optimal domino tilings cannot be neighbors.


Proof. Any two neighboring monominoes can be replaced with adomino, and so the original tiling cannot be optimal.

Theorem 113. All optimal tilings of a region have the same number ofblack, and the same number of white monominoes.

[Not referenced]

Proof. Suppose we have two tilings, T with W white monominoesand B black monominoes, and T′ with W ′ and B′ white and blackdominoes respectively, and let S be the subregion covered by domi-noes in T, and S′ be the subregion covered by dominoes in T′.

Now we have ∆(R) = ∆(S) + B −W = ∆(S′) + B′ −W ′. But Sand S′ are tileable by dominoes, so ∆(S) = ∆(S′) = 0 (Theorem 23).So B−W = B′ −W ′. But also, since the tilings are minimal, we haveB + W = B′ + W ′, which gives is B = B′ and W = W ′.

Class G(R)

Strip polyomino with odd area (includes snake, rectangle) 1Column convex polyomino with k odd columns (includes cylinder) ≤ kBar graph See Bodini and Lumbroso (2009)Young diagram (includes L-shaped polyomino, rectangle) |∆(R)|

Table 11: Gap numbers for some classesof polyominoes.Theorem 114. The gap number of a Young diagram R is |∆(R)|.

[Not referenced]

Proof. Suppose we cannot remove a cylinder from the region. Then,the Young diagram must have top cells in every column the samecolor (Theorem 40). If we put a monomino on each of the k top cellsof odd columns, the remainder can be tiled with dominos. Clearly|∆(R)| = k, so the gap number must be at least k (Theorem 108) andsince we can find a tiling with k monominoes and dominoes, the gapnumber is exactly k.

Suppose we can remove a cylinder. We can continue this until wecannot. We now have the shape described above. Find a tiling for

108 polyominoes

it, and note that the absolute value of the deficiency is equal to thenumber of monominoes. It remains to show the original figure can betiled with the required monominoes.

Now re-insert cylinders that were remove (using the process ofTheorem 37) to find a tiling of the original region k monominoes.This means G(R) ≤ k. But G(R) ≥ |∆(R)| = k, therefor G(R) = k =

|∆(R)|.

Once we have determined the gap number, we can also ask: whereare legal positions for these monominoes to occur?

Example 13. In a rectangle with odd area, the deficiency is 1, and say WLGthat B > W. Then at the very least the monomino must be placed on a blacksquare. It is easy to see that a monomino can occur on the corner of such arectangle. Using this, we can show it is possible for this monomino to occuron any black cell.

First, suppose we want to place a monomino somewhere on border of therectangle. We can now divide the rectangle into two rectangles such thatone is even, and the other is odd with the monomino in the corner. Note thatwe can only do this when the monomino is placed on a black square, as isrequired. Both these are tileable, and so is the whole.

Suppose then the monomino is somewhere in the interior of the rectangle.Again, we can divide the rectangle into two rectangles, one which is even,and one which is odd with the monomino on the border. And again we canonly do this because the monomino has to be placed on a black square. Boththese are tileable, and therefor so is the whole.

Table 12 tabulates the number of tilings for some rectangles with odd areaby dominoes and a single monomino.

(a) A rectangle withodd area and amonomino placedin the corner is tileable.

(b) A rectangle withodd area and amonomino placedon a black cell the edgeis tileable.

(c) A rectangle withodd area and amonomino placedon a black cell in theinterior is tileable.

(d) A rectangle with amonomino on a whitecell is not tileable, sincethe remainder of theregion is unbalanced.

It is not always the case that monominoes can occur on any cell ofthe correct color. In Figure 111 for example, there is no optimal tilingof the region with the monomino on the red cell. In this case remov-ing the red cell partitions the remaining region into two regions withan odd area each, so neither is tileable by dominoes (Theorem 1). The

dominoes ii 109

same applies if the cell would partition the region into two regionsand one is unbalanced, as shown in Figure 112.

Figure 111: There is no optimal tiling ofthis region with the monomino coveringthe red cell.

Figure 112: The gap number of thisregion is 1, but there is no optimal tilingwith the monomino covering the redcell.

Figure 113: The optimal tilings of a3× 3 square.

1 3 5 7 9

1 1 2 3 4 5

3 2 18 106 540 2554

5 3 106 2180 37924 608143

7 4 540 37924 2200776 116821828

9 5 2554 608143 116821828

Table 12: Number of tilings of odd-arearectangles with one monomino anddominoes.

The following Theorem generalizes Theorem 61.

Theorem 115. Suppose R has two tilings T and U by dominoes andmonominos. Then for each monomino in T that is not in U, there is a stripS that is tiled by a set of dominoes and a monomino in each tiling, and themonomino is at one of the strips ends in each tiling.


Proof. Let C1 be a cell tiled by the monomino in T. That cell is cov-ered by a domino in U, and has neighbor in that domino C2. C2 is

110 polyominoes

covered by a domino in T, it’s neighbor is C3. We can continue thisprocess until we reach Ck, which covers a monomino in U.

Note the following:

• k must be odd (this can be seen from the process itself, or fromthe fact that the two monominoes must have the same color,and hence any strip of which they are the endpoints must havean odd number of cells).

• Ci 6= Cj if i 6= j. For suppose Ci = Cj and i < j. It followsthat Ci−1 = Cj−1, and so on until we have C1 = Cj−i+1. Butunless i = j, it means that Cj−i+1 is not covered by a dominoin T (it is covered by a monomino), and therefor cannot havebeen part of the sequence in the first place, because all cells inthe strip Ci 6=1 are covered by dominoes in T.

Suppose that T is a tiling of R, and that C1, · · ·Ck ∈ R is a strip,and that C1 is covered by a domino in T, and C2i, C2i+1 is covered bya single domino in T. We can now form a new tiling T′ by coveringeach C2i−1, C2i by a domino, and Ck by a monomino. This transfor-mation is called a strip shift. See Figure 114 for an example.

Figure 114: An example of a strip shift.

The following Theorem generalizes Theorem 63.

Theorem 116. Any tiling of a region R by dominoes and k monominoescan be transformed into any other tiling with dominoes and k monominoesby performing strip rotations and strip shifts.


Proof. Let T and U be two tilings of a R. For each monomino in T notin U, there is a strip Si such that if we transform a strip shift in T, weget the strip’s tiling in U (Theorem 115). Let Mi be the cells coveredby monominoes in both T and U. Then R− ⋃i Si −

⋃i Mi is a region

with subtilings in T and U, that is covered by only dominoes in bothtilings. These tilings are connected by strip rotations (Theorem 63).

Here are some consequences of Theorems 115 and 116:

(1) In an optimal tiling, no two monominoes can lie in the sametiled strip. This is a generalization of Theorem 112.

(2) A region R cannot be partitioned into strips so that fewer thanG(R) of them have odd length.

dominoes ii 111

4.3.2 Tatami Tilings

9 A tatami tiling is a tiling such that no four corners of tiles meet 9 This is a placeholder section which Ihope to expand in a future version.(Knuth (2009, Problem 7.1.4.215); see also Ruskey and Woodcock

(2009) and Hickerson (2002)). Notice that this definition does notreference any tileset. There are two types of tatami tilings that areusually considered: tatami tilings with dominoes, and tatami tilingswith dominoes and monominoes.

Figure 115: Tatami tilings of the 4× 4square with dominoes and monomi-noes.

The plane has many tatami tilings with dominoes; several areshown in Figure 116.

(a) Horizontal Bond Pattern (b) Vertical Bond Pattern (c) Herringbone Pattern

Figure 116: Tatami tilings of the plane.

4.3.3 Fault-free tilings

A fault in a tiling is a line on the grid (horizontal and vertical) thatgoes through the region and is not crossed by any tile (Golomb, 1996,p. 18).10 See Figure 117 for an example. A tiling with no faults are 10 Also called line of cleavage, Reid (e.g.

2005). The problem of finding fault-freetilings was first proposed by Robert I.Jewett according to Golomb.

called fault-free. We are mostly interested in fault-free tilings ofrectangles. See Figures 118 and 119 for examples.

Figure 117: A tiling with a horizontalfault.

Theorem 117 (Golomb (1996), p.18). Domino tilings of 2× n rectanglescannot be fault-free.


Proof. There are only two configurations in which a rectangle canstart (Figure 101). Both of these have faults.

112 polyominoes

Theorem 118 (Golomb (1996), p.18 (no proof)). Domino tilings of 3× nrectangles cannot be fault-free.


Proof. A rectangle can only start with a universal block, a tower,or one of the start blocks (see Figure 102). The universal block andtowers all have faults. A start block can only be followed by an endblock, or a mid-block; and a mid-block can only be followed by an-other mid-block or an end block. So, since the rectangle is finite, if itstarts with a start block, it must eventually have an end block. If thatis the final block, the rectangle has a horizontal fault. Otherwise, ithas a vertical fault.

Theorem 119 (Golomb (1996), p.18). Domino tilings of 4× n rectanglescannot be fault-free.


Proof. (Martin, 1991, p. 18) All the ways a 4× n rectangle can beginis shown in Figure 103. The first 5 can also end the rectangle, or theycan be extended. In the former case, each of those 5 has a fault. Inthe later case, there is a vertical fault at the end of the block.

The third last can be extended only by adding horizontal domi-noes to the top and last row to the right. This shape can either becompleted with a vertical domino, in which case there are two hor-izontal faults, or with two horizontal dominoes, which gives us thesame situation as we started with (so eventually, a fault must be in-troduced).

The final two shapes can be extended by either a vertical domino,or two horizontal dominoes. In the former case, we have a horizontalfault, or if the rectangle continues, a vertical fault. The the latter case,we have the same situatiom we started with, so eventually a faultmust be introduced.

Theorem 120 (Golomb (1996), p.18). The 5× 6 rectangle has a fault-freetiling by dominoes.


Theorem 121 (Golomb (1996), p.18). The 6× 6 has no fault-free tiling bydominoes.


dominoes ii 113

Proof.Figure 118: The two fault-free tilings ofa 6× 5 rectangle.

Proof. Consider any potential fault line. This line must be coveredby a domino. If no other domino covers it, then the line divides thefigure into two figures of odd area each (the area is odd since thecells covered by the fault-covering domino are not included). Thenthere must be at least one other domino on the line (Theorem 21).Since there are 10 lines, we need 20 dominoes. But the area is only36, which means it will be covered by 18 dominoes. It is thereforeimpossible to cover all faults and still tile the figure. Thus, the figurehas no fault-free tiling.

Theorem 122 (Golomb (1996), p.19). A 6× 8 rectangle has a fault-freetiling.


Proof.Figure 119: A fault-free tiling of a 6× 8rectangle.

Theorem 123 (Golomb (1996), p.19). If we have a fault-free tiling of aR(m, n), we can find fault-free tilings of R(m + 2, n) and R(m, n + 2).


Proof. The same argument works for both types of extensions, solet’s prove it for horizontal extensions. Pick a vertical domino onthe border. (Such a domino must exist, for if it does not, then therewould be a fault.) Remove it from the tiling. Now add a vertical2-cylinder of dominoes along the crooked edge. Add the removeddomino in the opening. The new rectangle is 2 cells wider, and isstill fault-free. To see this, note that all original fault lines are still

114 polyominoes

covered. We introduced two new potential faults. The inner mostis covered by the dominoes that fit into the opening of the crookededge; the other is covered by the remaining dominoes.

Figure 120: Extending a fault-free tilingof a 6× 5 rectangle horizontally andvertically.

Theorem 124 (Golomb (1996), p.18). All even-area rectangles R(m, n)(except R(6, 6)) has fault-free tilings by dominoes if m, n ≥ 5.

[Not referenced]

Proof. By Theorem 123, we can use suitable extensions to find fault-free rectangles for R(5 + 2p, 6 + 2q) and R(6 + 2p, 8 + 2q) , since both5× 6 and 6× 8 rectangles as has fault-free tilings (Theorems 120 and122). Clearly a 1× m rectangle must have faults. Theorems 117, 118,and 119 shows we cannot have fault-free tilings when 1 < m ≤ 4 or1 < n ≤ 4. And finally, Theorem 121 shows there are no fault-freetilings of the 6× 6 rectangle.

Figure 121: Fault-free tiling of theplane.

Problem 34.

(1) What rectangles have fault-free tatami tilings with

dominoes ii 115

(a) dominoes only,

(b) dominoes and monominoes?

(2) Prove any tiling of an Aztec diamonds has at least three faults.

4.3.4 Even Polyominoes

An even polyomino is a polyomino where all corners (of inside andoutside borders) neighbor black cells if we color the polyomino andits surroundings with the checkerboard coloring (Kenyon, 2000b).Note that the holes of an even polyomino are even polyominoesthemselves.

(a) (b)Figure 122: Valid corners in evenpolyominoes.

(a) (b) (c)

Figure 123: Examples of even polyomi-noes.

Theorem 125. Let R be a even polyomino with H holes. Then

∆(R) = 1− H. (4.12)


Proof. For each edge Ei, notice that:

(1) if Ei is a mountain, it lies between two black cells, and henceb(Ei)− w(Ei) = 1,

(2) if Ei is a valley, it lies between two white cells, and henceb(Ei)− w(Ei) = −1, and

(3) if Ei is a flat, it lies between a black and white cell, and henceb(Ei)− w(Ei) = 0.

Therefor, we have for the entire polyomino, b(R) − w(R) = P − V.By Theorem 27 we have ∆(R) = b(R)−w(R)

4 , so ∆(R) = P−V4 . And by

Theorem 13 we have P−V4 = 1− H, so ∆(R) = 1− H.

It follows immediately that even polyominoes without holes havean odd area. Also note that in even polyominoes, mountains andvalleys have odd length, and flats have even length.

Problem 35.

(1) Show that if we insert a 2-cylinder into an even polyomino, the resultis an even polyomino.

116 polyominoes

(2) If we remove a 2-cylinder from an even polyomino, is the result aneven polyomino?

Theorem 126. Even polyominoes with one hole are tileable.


Proof. For this proof to work, we will manipulate the border (eitherthe outside or inside border) to give a new polyomino with a hole,and we will consider regions where borders overlap (but not cross)also valid polyominoes, and they are even if the corners are black aswith the regular definition. Here are some examples of such poly-ominoes (the outside border has been slightly displaced where itcoincides with the inside border).

Figure 124: Border polyominoes.(1) In an even polyomino, all peaks must be black (otherwisethere are white corners). Also, all black peaks must be at-tached to a white cell with exactly two neighbors. Therefor,in an even polyomino, if we move the border to exclude thepeak and its neighboring cell, we are still left with an evenpolyomino.

An example of this manipulation is shown in Figure 125.

Figure 125: Peak removal(2) A 2× 2 square can be of two types. Type I has a black square

in the top right, a Type II square has a black square in the topleft. If a Type I square occurs in a corner of the polyomino, wecan move the border to exclude it, and keep the polyominoeven.

An example of this manipulation:

Figure 126: Corner removal

(3) All convex corners of a even polyomino are one of the fourtypes shown in Figure 127.

(a)Type1

(b) Type2

(c) Type3

(d) Type4

Figure 127: Corner Types

(4) If we perform the manipulations in 1 and 2 until we cannot,we are only left with Type 1 and Type 2 corners.

(5) A region with only type 3 and 4 convex corners and one holecannot contain a 2× 2 subregion.

dominoes ii 117

Suppose there is such a square. Find the largest connected setof cells that contains this square such that each cell is part of a2× 2 subregion. This set of cells must have at least four convexcorners (Theorem 8). For these corners to not be corners of theentire region, each needs to be “covered” by a “strand” (thepicture below shows a possibility.) These strands must eitherresult in peaks, or they must connect in pairs. If they connectin pairs, there are two holes, and this is impossible (Figure128). Therefor, there cannot be any 2× 2 subregion.

Figure 128: An example where strandsthat cover potential corners meet up inpairs, leading to two holes.

In this image, there is a 2 × 2 square, contained in a 3 × 3square where each corner is covered by a white cell that starsthe strand. In this example, the strands connect in pairs, lead-ing to two holes.

(6) We cannot have any X- or T-junctions in a region with no 2× 2subregions, no peaks, and one hole.

Suppose we have an X-junction. Each of the four connectorsmust eventually lead to a peak, or pairs of them must join.In the latter case, we have two holes, which is impossible.Therefor there are no X junctions.

Suppose we have a T-junction. Each of the three connectorsmust either lead to a peak, or they must join other connectors.We can only join up if we have at least another T junction,in which case we will have at least have two holes, which isimpossible. Therefor there are no T-junctions.

(7) This means we either have a ring, or a polyomino with no area(thus the inner and outer border coincide completely).

(8) This procedure shows we can partition any even polyominowith one hole into a set with dominoes, 2 × 2 squares, andeither a ring (tileable by Theorem 51) or empty polyomino. Allthe elements of this partition are tileable, and therefor, so is thewhole region (Theorem 2).

The double border definition is necessary to deal with polyomi-noes such shown in Figure 129. If we did not have that, removing a

118 polyominoes

2× 2 corner would not leave an even polyomino. (It would be nice tofind a proof that does not require this trick.)

Figure 129: An example of a polyominofor which our proof would fail if we didnot use the double border definition ofa polyomino.

One consequence of using this formulation is that the hole neednot be an actual hole using the normal definition of a polyomino.

An algorithm follows immediately from the border manipulations.

Theorem 127. A rectangle with odd area with an odd hole is tileable if it isbalanced.

[Not referenced]

Proof. Let R = R1 − R2 be our region, and R1 the filled rectangle,and R2 the hole. The deficiency of an odd-area rectangle is 1 or -1.(You can pair rows to cancel out, leaving the last row with one morecell of one color than the other.) Suppose the corners of the rectangleis black. Then, for the region to be balanced, the corners of the holemust also be black, because we have ∆(R) = ∆(R1) − ∆(R2), or0 = 1 − (B −W), So B −W = 1, and so we have B > W, whichmeans the rectangle must have black corners. Therefor R is an evenpolyomino, and so is tileable by dominoes (Theorem 126).

We can “fix” the deficiency of even polyominoes in a specific wayto make a class of polyominoes that are tileable by dominoes.

A Temperleyan polyomino is a polyomino formed from an evenpolyomino by removing one black cell from the outer border, andappending a black cell to each inner border Kenyon (2000b).

Theorem 128. Temperleyan polyominoes are tileable by dominoes.


Proof. Case h = 0. Suppose R is the polyomino, and H is the blackcell removed from the border. Then let R′ = R ∪ H. We can now viewthis polyomino as a double border polyomino, with outer border theborder of R′, and the inner border the border of H. Since R′ is evenand H is an even hole, R must be tileable by Theorem 126.

General case. See Kenyon (2000b, Section 7) for a sketch of a proof.

dominoes ii 119

Example 14. Suppose R is a region partitioned into two simply-connectedsubregions S1 and S2, with S1 a black even polyomino and S2 a white evenpolyomino. Then R is tileable. Let v1 ∈ S1 and v2 ∈ S2 be neighbors withv1 black (and v2 white), and let S′1 = S1 − {v1} and S′2 = S2 − {v2}. ThenS′1 and S′2 are both Temperleyan polyominoes, and tileable. And {v1, v2} istileable by a single domino since v1 and v2 are neighbors. The three regionsform a partition of R, and are all tileable, and therefor R is tileable (Theorem2).

Theorem 129. The gap number of an even polyomino with H holes is|H − 1|.

[Not referenced]

Proof. Case H = 0. If we remove a black corner cell, the resultingpolyomino is Temperleyan and thus tileable by Theorem 128. There-for, the gap number is 1.

Case H = 1. The region is tileable (Theorem 126), and so the gapnumber is 0.

Case H > 1. Form a Temperleyan polyomino R′ by appendingblack cells vi to all but one interior border; there are H − 1 such cells.This polyomino is tileable (Theorem 128). Let ui be the neighbor ofvi in the same domino of a tiling of R′, and let R′′ = R− {ui}. ThenR′′ is tileable, and R therefor has a tiling by dominoes and H − 1monominoes. Since g ≥ |∆(R)| (Theorem 108), and ∆(R) = 1− H(Theorem 125), the tiling is optimal, and therefor g = |1− H|.

4.4 Summary of tiling criteria

4.5 Miscellaneous

4.5.1 Polyominoes that can be Tiled by Dominoes

How many regions of a given area is tileable by dominoes? Table14 gives us a sense. A polyomino tileable by dominoes is called apolydomino.

Figure 130: The four tileable tetromi-noes.

120 polyominoes

Conditions for tileability Theorem

Necessary

|R| must be even. 1

R is balanced. 23

The parity of dominoes that cross the border of any subregion S of R mustequal the parity of the deficiency of S.

21

The deficiency of any subregion must match the flow (and therefor, thenumber border of S must be able to accommodate the necessary flow.

25

R must be have least 2 sides of even length. 29

R S is tileable if S is an n-cylinder with n even, and removing S from R issafe.

38

R is fair by some discriminating coloring. 47

Sufficient

All sides of R is even. 30

R S is tileable, with S a n-cylinder with n even 37

R is a strip-polyomino with even area, which includes rectangles with evenarea.

51

R is saturnian 55

R is a checkerboard-colored strip polyomino with two cells of oppositecolors removed

56

R is a balanced stack polyomino (which includes balanced Young dia-grams).

46 (41)

R is a balanced jig-saw region. 45

R is an even polyomino with one hole. 126

Necessary and Sufficient

It has no bad patches 34

R is fair by all discriminating colorings 49

We can consistently assign heights in each step of Thurston’s Algorithm. 92

R is simply connected, balanced, and the difference in height is smaller thanthe distance between any two vertices on the border.

95

Table 13: Summary of tiling criteria fordomino tilings.

dominoes ii 121

n 2n-ominoes Balanced Tileable Uniquely tileableA210996 A234012 A056785 A213377

1 1 1 1 1

2 5 4 4 3

3 35 24 23 20

4 369 230 211 170

5 4655 2601 2227 1728

6 63600 32810 25824 18878

7 901971 433855 310242 214278

8 13079255 5923677 3818983 2488176

9 192622052 47752136 29356463

α 14.7* 13.6* 12.5* 11.8*

Table 14: Balanced, tileable anduniquely tileable dominoes.

Figure 131: The 23 tileable hexominoes.All except the last three have uniquetilings.





122 polyominoes

4.6 Further Reading

Berlov and Kokhas (2004) give some interesting problems (with solu-tions) on domino tilings.

For more on orders and lattices, see for example Davey and Priest-ley (2002) and Roman (2008). The following properties of lattices arewell-known and it is worth considering their implication on dominotilings:

(1) M3 is not isomorphic to any sublattice. (Davey and Priestley,2002, Thereom 4.10, p. 89)

(2) N5 is not isomorphic to any sublattice. (Davey and Priestley,2002, Thereom 4.10, p. 89)

(3) The covering relation ≺ forms a median graph. (Roman, 2008,Ex. 9, p. 125)

For more on the lattice structure of domino tilings specifically, seeRémila (2004), Caspard et al. (2003). For slightly more general resultson height functions and dominoes tilings see Ito (1996), Beauquierand Fournier (2002) and Bodini and Latapy (2003).

Domino tilings (especially the use of height functions and thestructure of tilings) have also been considered for regions more gen-eral than was covered here. For example:

• plane regions with barriers (Propp and Stanley, 1999)• three-dimensional regions (Milet, 2015)• regions on a torus (Impellizieri, 2016)• quadriculated surfaces (Saldanha et al. (1995) and Saldanha

and Tomei (2003))

Height functions and lattices can also be used to study other tilingproblems11, for example, tilings with: 11 I use here the convenient list provided

by Kenyon.• bars (Kenyon and Kenyon, 1992)• rectangles (Kenyon and Kenyon, 1992)• squares (Kenyon, 1993)• tetrominoes (Muchnik and Pak, 1999)• bibones (Kenyon and Rémila, 1996)• tribones (Thurston, 1990)• lozenges (Rémila (1996) and Thurston (1990))• leaning dominoes (Rémila, 1996)• triangles (Rémila, 1996)

In our treatment of height functions, we did not deal with regionswith holes. The best explanation for dealing with holes is Bodini andFernique (2006), but it is covered in the slightly more general setting

dominoes ii 123

of planar dimer tilings. Desreux et al. (2004) gives a similar techniquespecific for domino tilings, including algorithms for finding mini-mum tilings, to generate all tilings, and to do uniform sampling oftilings. Saldanha et al. (1995) gives an alternative approach, but re-quires some ideas from topology. Thiant (2003) gives an algorithm fortiling any region with dominoes.

In many of the proofs I omitted the details of solving recurrencerelations. How to solve these is explained in many books on com-binatorics or discrete mathematics; see for example (Brualdi, 2009,Chapter 7), (Knuth et al., 1989, Chapters 1 and 7) and (Wilf, 2005,Chapter 1).

Propp and Lowell (2015) discusses enumeration of tilings in gen-eral, and is a good overview of some of the techniques used. Mathe-maticians calculated the number of domino tilings of various classes,for example:

• quartered Aztec diamond (Lai, 2014)• deficient Aztec rectangle (Krattenthaler, 2000)• holy square (Tauraso, 2004)• L-shaped domain (Colomo et al., 2018)• expanded Aztec diamond Oh (2018)

Counting more general types of matchings is discussed in Proppet al. (1999). Special attention has been given to matchings whosecounts are perfect powers (such as the Aztec diamond). See for exam-ple Pachter (1997), Ciucu (2003) and Lai (2013). One notable result isa factorization theorem for graphs with reflective symmetry given inCiucu (1997).

Numerical results and generating functions for various tilingsof rectangles by small rectangles (including dominoes) is given inMathar (2013) and Mathar (2014). This includes counts of tatamitilings. In addition to generating functions, Gershon (2015) givesexact formulas for rectangles of fixed width up to 6.

For more details on tatami tilings with dominoes only, see Alhazovet al. (2010) and Ruskey and Woodcock (2009). For more on tatamitilings with dominoes and monominoes, see Erickson (2013), Ericksonet al. (2011) and Erickson and Schurch (2012). The idea of water strid-ers in Erickson et al. (2011) is closely related to Theorems 98, 97, and96.

We consider optimal tilings of rectangles by rectangles and theassociated gap number in Section 6.1.2. Fault-free tilings of otherpolyominoes is covered in Section 6.4. Temperleyan dominoes areclosely related to spanning trees of certain graphs. This is discussedin Kenyon et al. (2001), with applications to the dimer model inKenyon (2000b), Kenyon (2001) and Kenyon (2000a).

5Various Topics

The sections in this chapter can be considered “short chapters”; theyare independent topics that introduce material useful for later chap-ters, but without enough material to warrant their own chapters.

5.1 The Tiling Hierarchy

Figure 132: The relationship betweenclasses. Classes in blue blocks haveelements that are not in any higherclass; white ones have no examplesof polyominoes that are not also in ahigher class.

Up to this point, we have looked mostly of tilings of arbitrary re-gions. In this section, we look at tilings of some “fundemental” re-gions, most of which are not finite. We will also see that the regionsconsidered in this section forms a hierarchy; if a polyomino set tilesone of these regions, it tiles all the regions below it in the hierarchy.

These regions form a hierarchy (as established in Theorem 133),so that when we know something about whether a polyomino tiles acertain class, we also have information about whether it can tile someof the other classes.

The tiling hierarchy is the classes defined in Golomb (1966) andextended in Liu (2018a). We also split the original Plane class intotwo (aperiodic and periodic).

A reptile is a polyomino that can tile a scaled copy of itself.Monominoes tile everything, including bigger squares, therefor

monominoes are reptiles. So is any rectangle. The first interestingreptile is the straight tromino. The skew tetromino is the smallestpolyomino that is not a reptile. The T-tetromino is a reptile — but ina boring way: it tiles a square, and so we can use the square to tile abigger T-tetromino. Of course this is a general idea: if a polyominotiles a rectangle, it can tile a square, and if it tiles a square, it cantile any scaled polyomino, including itself. Although the converse isnot proven to be true, it is an open problem (Winslow, 2018, OpenProblem 13)]; all reptiles we know also tile rectangles.

That reptiles tile the plane should be clear: take the polyomino,dissect it in smaller polyominoes of the same shape. Scale the figureso that the polyominoes in the dissection are the same size as the

various topics 125

Class Can Tile

N NothingAP Plane (aperiodic)PP Plane (periodic), torusHP Half planeQ QuadrantS Strip, CylinderXS Crossed StripTS Branched StripHS Half-stripBS Bent-stripR RectangleI Scaled copy of itself

Table 15: Tiling Classes. For example, ifsome polyomino P ∈ R it means it cantile a rectangle.

original. Repeat. This way, any size patch can be covered. (This typeof tiling is called a substitution tiling.)

What is perhaps surprising is that reptiles always tile a quadrant.We prove this in the next few theorems.

Figure 133: The T-tetromino is a reptile.Theorem 130 (Golomb (1966), Theorem 4). A reptile covers at least onecorner of its rectangular hull.


Proof. Make a big version of the reptile by the replication process,(as big as you want). Clearly, if the reptile does not cover all fourcorners, there is no why it can cover the corner of the bigger copy;there will always be a gap.

A reptile can fail to cover two corners; 134 shows an example.

Figure 134: An example of a reptile thatcovers only one corner of its rectangularhull.

Theorem 131. A polyomino that can tile a rectangle, can also tile a square.


Proof. Suppose the polyomino tiles R(m, n). You can stack n of theserectangles together to tile R(mn, n), and then stack m of these to tileR(mn, mn), which is a square.

Theorem 132. A polyomino that tiles a rectangle tiles can tile any poly-omino scaled by a factor.

[Not referenced]

Proof. By Theorem 131 a polyomino that tiles a rectangle can tile asquare. These squares can be stacked to construct the scaled poly-omino.

126 polyominoes

The constructions in the proofs of the two theorems above doesnot necessarily construct the smallest possibility. For example, a 2× 4polyomino can also tile a 4× 4 square.

Theorem 133 (Golomb (1966)). For a single polyomino:

BS ⊂ Q ⊂ HP (5.1)

R ⊂ I ⊂ Q ⊂ HP (5.2)

R ⊂ HS ⊂ BS ⊂ TS ⊂ XS ⊂ S ⊂ HP ⊂ PP ⊂ N (5.3)


Proof. All the inclusions are obvious except for the following two:

XS ⊂ S (5.4)

I ⊂ Q (5.5)

To prove 5.4, let P tile a crossed strip. Let m be the maximum di-mension of the polyomino’s rectangular hull, and let w be the widthof one arm of the crossed strip. Now in any w× m rectangle of thatarm, we can draw at least one path from the top of the arm to thebottom. There are only finitely many of such patterns, so in the armwith infinitely many such rectangles, at least one pattern must repeat.The segment of the arm between such repeated patterns is a cylinder,and it can used to tile a strip by placing them end to end.

(a) Half strips tile abent strip.

(b) Bent strips tile abranched strip.

(c) Branched stripstile a crossed strip.

(d) Bent strips tile aquadrant.

(e) Quadrants tile ahalf plane.

(f) Half planes tilethe plane.

Figure 135: Some inclusions of the tilinghierarchy demonstrated.

To prove 5.5, let P be a reptile. It must cover a corner of its rectan-gular hull (by Theorem ). Place the polyomino in the first quadrantwith this corner at the origin. Now perform the replication processuntil the covered corner of the big polyomino is covered by a smallpolyomino in the same orientation. (This process can go on for atmost 8 iterations, since there are only 8 distinct orientations .) Movethis bigger polyomino to the origin, and call the new figure F(P). Byrepeatedly applying F, we can expand the tiling, which will cover theentire quadrant in the limit.

While we often use this theorem in the "forward" direction, it isworth considering reverse implications. For example, if a polyominodoes not tile a quadrant, it can also not tile a rectangle. The followingexample shows how knowledge about how a tile tiles the quadrantinforms us about how it tiles a rectangle.

Example 15 (Reid (2003), Example 6.7). The polyomino B(2 · 3 · 22) tilesR(m, n) iff 6 | m and 6 | n.

Proof. First, note that of all 8 possible placements along an edge, only4 are possible. Second, by examining all cases (not hard to do) weobserve that the edge must always be made of pairs as shown below.

various topics 127

Further examination shows the only way to tile the quadrant is touse 6× 6 squares. And since no other tiling of the quadrant exists, itimplies all rectangles must be tiles using these squares too, which isonly possible if 6 | m and 6 | n.

Figure 136: These polyominoes canonly pack an edge in the configurationshown.

No-one knows whether the following statements are true (Winslow(2018, Open Problems 12, 13), Golomb (1996, p. 107–108).)

(1) I = R

(2) HS = R

(3) Q = BS (and indeed, if I 6= R, then whether Q = I∪BS).

(4) HP = S

(5) PP = AP

There is no reason to believe any of these statements, other thanthat no counter examples have been found. There are suggestionsthat these may not be true:

• The statements (2)-(5) do not hold for tilesets that have morethan one polyomino. For example, .

• The Y pentomino tile half-strips of width 6 and 8, bit not anyrectangles with sides of length 6 or 8. This suggests HS 6= R.

• The P pentomino is an example of a reptile that does not re-quire construction of a rectangle first.

All polyominoes with area 6 or less can (at least) tile the plane.Table 16 lists the highest classes that polyominoes belong to.

Problem 36. Is the following true or false: If a polyomino tiles anotherpolyomino with a scaling factor of n > k, where k is the largest side of therectangular hull of the polyomino, it tiles a rectangle.

Problem 37. Complete the Table 16.

128 polyominoes

Polyomino Class

Monomino R

Domino R

Bar R

Right R

Bar R

Square R

T R

L R

Skew BS

I R

L R

P R

Y R

N BS

W BS

F TS

V TS

T PP

U PP

X PP

Z PP

R

R

R

R

R

R

R

R

R

R

Polyomino Class

BS

BS

BS

BS

BS

BS

TS

TS

TS

XS

XS

XS

S

S

S

PP

PP

PP

PP

PP

PP

PP

PP

PP

PP

Polyomino Class

0 R

PP

PP*, Q′

PP

PP*, Q′

PP*R

PP

N*S*R

N*

N*N*N*N*N*N*N*N*N*

N*N*N*N*N*

Polyomino Class

N*N*

N*

N*

Table 16: Tiling classes for polyominoes.The table gives the smallest class apolyomino is part of. Classes markedwith an asterisk is not guaranteed tobe the smallest. An accent shows thepolyomino is not in that class.

various topics 129

5.1.1 Strips, Crossed Strips, Bent Strips, and Half Strips

We make the following set of definitions:

• A strip is prime if no smaller strip tiles it.

• A half strip is prime if no smaller half strip tiles it.

• A bent strip is prime if no smaller bent strip tiles it.

• A rectangle is prime if no smaller rectangle tiles it.

Similar definitions hold for larger tile sets.From prime regions of a tile set, we can construct all regions of

the same type that is tileable by the set. We can also, to some extent,given an arbitrary region, answer whether it is tileable by a tile set ifwe know all the prime region of the relevant type. The case for rect-angles is interesting, and treated in detail in the chapter Rectangles.

For strips and half strips, finding all strips or half strips is equiv-alent to the Frobenius problem.The prime strips and halfstrips aregiven in table 17. A rectangle R(m, n) can be stacked to form a stripof widths m or n, but not all strips correspond to prime rectangles.Those that don’t are marked in the table.

(a) HS(6) (b) HS(8)

Figure 137: Halfstrips of the Y pen-tomino that does not correspond towidths of prime rectangles.

Theorem 134. If a tile set tiles a strip, there exists a periodic tiling by thesame set.

130 polyominoes

Polyomino Prime Half Strips Prime Strips

Monomino 1 1

Domino 1 1

Bar 1

Right 2 3 2 3

Bar 1 1

Square 2 2

L 2 3 2 3

T 4,...? 2, 13d

F - -I 1 1

L 2 5 2 5

N - -P 2 3 2 3

T - -U - -V - 5 ...?W 5 ...?X - -Y 5, 6

a, 8a, 9 2

b, 5

Z - -

Yc12

a, 23, 29, 30, 32 2, ...?

Table 17: Prime Half Strips and Strips.aNo rectangles with this width.bNo half strips with this width.cPrime ?dHochberg (2015)

various topics 131

[Not referenced]

Proof. Lets suppose we have a tiling of a strip by a set of polyomi-noes.

Now make new tiles by taking each column as a tile. Tiles arethe same if they cut though the original polyominoes in exactly thesame way. This new tile set is finite. Now construct a directed graphwith the new tiles as vertices, with an edge from Ti to Tj if Tj can beplaced on the right of Ti and is legal by considering the original set(that is, all the polyominoes formed are from the original set).

The tiling if the strip corresponds to an infinite path along thedirected edges. But since there is a finite number of vertices, theremust be a cycle in the graph. This cycle corresponds to a periodictiling with the tiles, and so also a periodic tiling with the originaltiles.

A cycle in the proof above corresponds to a cylinder. The theoremis then equivalent to the following:

Theorem 135. If a tile tiles a strip of width m, it also tiles a cylinder ofwidth m.

[Not referenced]

Problem 38. Complete Table 17.

5.1.2 Extending the hierarchy

It is not hard to come up with various other classes that fits into theexisting hierarchy. It is much harder to tell whether such extensionswould be useful. In this section, I give some ideas for classes thatmight be.

RR Tiles a rectangle with a rectangular hole. It is easy to see thatR ⊂ RR. The T- and V-pentominoes are examples of polyomi-noes in RR but not in R.

Rectangle with n rectangular holes

Rectangle with an arbitrary connected hole

Rectangle with n arbitrary holes

Corner

Edge

Half Edge

Open Edge

132 polyominoes

The plane with holes

The more classes there are, the more tedious it is to find the char-acteristic class of a polyomino or set of polyominoes. In the tablebelow we only classify polyominoes up to pentominoes.

The notion of prime shape also becomes more complicated, al-though it may be interesting to explore.

5.1.3 Further Reading

Reid (1997) give several examples of families of polyominoes that tilestrips, but it is unknown whether they tile any rectangles.

Nitica (2018a) and Nitica (2018b) explores how the tiling hierarchyis affected when we are only allowed to tile by translation. Otherclasses of regions is also considered.

5.2 Colorings

If we color the plane with k colors, how many colors do we need sothat a polyomino does not cover two cells of the same color no matterhow it is placed?

If we need k colors for a polyomino P, we call this the chromaticnumber of P and denote it by C(P). We will use C(m, n) as an abbre-viation for C(R(m, n)). A coloring of the plane with C(P) colors suchthat however we place P, it does not cover two cells of the same coloris called an sufficient coloring for P. A sufficient colorings that mini-mize the number of colors — and therefor use k colors — is called anoptimal coloring for P;

We use the following notation for some common colorings. Thefirst is a generalization of the checkerboard coloring, which we call aflag coloring.

Fm,n(x, y) = (x + yn) mod m. (5.6)

In this coloring, there are m colors, each repeated in every row(in the same order), and each row offset by n cells from the previousrow.

If n = 1, we simply write Fm. In this notation, the checkerboardcoloring is F2.

The second is called a square coloring.

Sm(x, y) = (x mod m) + (y mod m)m. (5.7)

And this is simply the grid divided into m × m squares, eachcontaining m2 colors in the same pattern.

various topics 133

The final type of coloring is their product:

Sk × Fm,n(x, y) = Fm,n(bx/kc , by/kc)k2 + Sk(x, y). (5.8)

Problem 39. Show that the number of colors for

(1) Fm,n is n,

(2) Sm is m2, and

(3) Sk × Fm,n is km.

Theorem 136 (Yu (2014)). For any polyomino P, C(P) ≥ |P|.


Proof. Obvious.

Polyominoes for which C(P) = |P| are called efficient. Otherwisethey are inefficient.

Theorem 137 (Yu (2014), Implicit). Suppose the hull of P is R(m, n).Then C(P) ≤ C(R(m, n)).


Proof. Suppose C(P) > C(m, n). Color the plane with an optimalcoloring for R(m, n). Since this coloring requires less than C(P) col-ors, some placement of P covers two cells of the same color. If we putR(m, n) to cover P in such a placement, then two covered by R(m, n)must have the same color, and therefor the coloring cannot be opti-mal for R(m, n), a contradiction. Therefor C(P) ≤ C(m, n).

If P can cover any two cells of a polyomino Q, we say P protectsQ. A polyomino that cannot protect a polyomino larger than itself iscalled weak. Bars and squares are weak.

Problem 40. Which rectangles are weak?

Theorem 138. All efficient polyominoes are weak.

[Not referenced]

The converse does not hold. For example, below we will see thatW-polyominoes with an odd number of cells are also weak, but theyare not efficient.

Theorem 139 (Yu (2014), Implicit). If P protects Q, then C(P) ≥ C(Q).


134 polyominoes

Proof. Suppose C(Q) > C(P). Color the plane with an optimalcoloring for P. This coloring requires less than C(Q) colors, someplacement of Q covers two cells of the same color. These cells can becovered by P, so the coloring cannot be optimal for P, a contradiction.Therefor, C(Q) ≤ C(P).

Theorem 140 (Yu (2014), Implicit). If Q ⊂ P then C(P) ≥ C(Q).


Proof. P can cover the whole of Q, and in particular any two cells ofQ. Therefor, by Theorem 139 C(P) ≥ C(Q).

Figure 138: The shape of Theorem 141:B(1m · (n− 1)).

Theorem 141. Let P = B(1m · (n − 1)). Then C(P) = C(m, n), withFm(m+1)−1,m(m+2) a sufficient coloring.


Proof. The hull of P is R(m, n), so C(P) ≤ C(m, n) (Theorem 137).P can cover any two cells of R(m, n), so C(P) ≥ C(m, n) (Theorem

139).Taking these together, we get C(P) = C(m, n).

Theorem 142. Suppose that P has a optimal coloring with n colors, and thecoloring is a square coloring. Then if we remove a cell from P and append itso that the removed cell has the same color to form Q, then C(Q) = C(P).

[Not referenced]

Problem 41. What are all the polyominoes which are equivalent under S3

to the A-hexomino (B(1 · 2 · 3))?

Theorem 143. If P fits into Q in all orientations, and Q tiles the plane bytranslation1, then C(P) ≤ |Q|. 1 A polyomino that tiles the plane by

translation is called an exact polyomino;exact polyominoes are discussed inSection 7.1.[Referenced on pages 134 and 141]

Theorem 144. If P ∈ All tiles the plane, then C(P) = |P|, and so P isefficient.


Proof. P fits in itself in all orientations, so by Theorem 143 we haveC(P) ≤ P. But by Theorem 136, we have C(P) ≥ P. Thus, C(P) =

|P|.

Theorem 145. uppose the period of the tilings of Theorems 143 or 144 isgiven by u = (m, 0) and v = (n, 1), then

various topics 135

[Not referenced]

The next few Theorems establishes what is known of R(m, n), animportant case since the rectangular hull establishes an upper boundfor the chromatic number of all polyominoes. One would imaginecalculating the chromatic number of rectangles is easy; however, theproblem is subtle and not completely solved.

nm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

2 2 4 8 8 12 12 16 16 20 20 24 24 28 28 32 32 36 36 40 40

3 3 8 9 15 18 18 24 27 27 33 36 36 42 45 45 51 54 54 60 63

4 4 8 15 16 24 32 32 32 40 48 48 48 56 64 64 64 72 80 80 80

5 5 12 18 24 25 35 45 50 50 50 60 70 75 75 75 85 95 100 100 100

6 6 12 18 32 35 36 48 60 72 72 72 72 84 96 108 108 108 108 120 132

7 7 16 24 32 45 48 49 63 77 91 98 98 98 98 112 126 140 147 147 147

8 8 16 27 32 50 60 63 64 80 96 112 128 128 128 128 128 144 160 176 192

9 9 20 27 40 50 72 77 80 81 99 117 135 153 162 162 162 162 162 180 198

10 10 20 33 48 50 72 91 96 99 100 120 140 160 180 200 200 200 200 200 200

11 11 24 36 48 60 72 98 112 117 120 121 143 165 187 209 231 242 242 242 242

12 12 24 36 48 70 72 98 128 135 140 143 144 168 192 216 240 264 288 288 288

13 13 28 42 56 75 84 98 128 153 160 165 168 169 195 221 247 273 299 325 338

14 14 28 45 64 75 96 98 128 162 180 187 192 195 196 224 252 280 308 336 364

15 15 32 45 64 75 108 112 128 162 200 209 216 221 224 225 255 285 315 345 375

16 16 32 51 64 85 108 126 128 162 200 231 240 247 252 255 256 288 320 352 384

17 17 36 54 72 95 108 140 144 162 200 242 264 273 280 285 288 289 323 357 391

18 18 36 54 80 100 108 147 160 162 200 242 288 299 308 315 320 323 324 360 396

19 19 40 60 80 100 120 147 176 180 200 242 288 325 336 345 352 357 360 361 399

20 20 40 63 80 100 132 147 192 198 200 242 288 338 364 375 384 391 396 399 400

Table 18: Computed values of C(m, n),assuming that colorings are periodic.Colored cells are values that have beenproven:

Theorem 147

Theorem 148

Theorem 152

Table 18 shows some values computed by computer. Only thosemarked are proved to be the lowest; I calculated the remainder of thevalues by exhaustively searching for the smallest periodic coloring. Itis not established that there exist an optimal coloring that is periodic.

Theorem 146. Assume m ≤ n. Then

C(m, n) ≤ m(2n−m).

[Not referenced]

Proof. All orientations of R(m, n) fit in B(nm, mn−m). Therefor C(m, n) ≤|B(nm, nn−m)| = nm + m(n−m) = m(2n−m).

Theorem 147 (Puleo (2018a)). Suppose m ≤ n. Then C(m, n) = mn ifand only if m divides n, and so

(1) C(m, 1) = m, with coloring Fm

(2) C(m, m) = m2, with coloring Sm.


136 polyominoes

2 3 4 5 6 7 8 9 10

2 S2 F8,3 S2 × F2 F12,5 S2 × F3 F16,7 S2 × F4 F20,9 S2 × F53 F8,3 S3 F15,4 S3 × F2 S3 × F2 F24,7 S3 × F3 S3 × F3 F33,104 S2 × F2 F15,4 S4 F24,5 S6 × F8,3 S4 × F2 S4 × F2 F40,9 S10 × F12,55 F12,5 S3 × F2 F24,5 S5 F35,6 F45,19 S5 × F2 S5 × F2 S5 × F26 S2 × F3 S3 × F2 S6 × F8,3 F35,6 S6 F48,7 S8 × F15,4 S9 × F8,3 S6 × F27 F16,7 F24,7 S4 × F2 F45,19 F48,7 S7 F63,8 F77,34 F91,278 S2 × F4 S3 × F3 S4 × F2 S5 × F2 S8 × F15,4 F63,8 S8 F80,9 S10 × F24,59 F20,9 S3 × F3 F40,9 S5 × F2 S9 × F8,3 F77,34 F80,9 S9 F99,10

10 S2 × F5 F33,10 S10 × F12,5 S5 × F2 S6 × F2 F91,27 S10 × F24,5 F99,10 S1011 F24,11 S3 × F4 S4 × F3 F60,11 S6 × F2 S7 × F2 F112,41 F117,53 F120,1112 S2 × F6 S3 × F4 S4 × F3 F70,29 S6 × F2 S7 × F2 S12 × F8,3 S12 × F15,4 S12 × F35,613 F28,13 F42,13 F56,13 S5 × F3 F84,13 S7 × F2 S8 × F2 F153,35 F160,4914 S2 × F7 S3 × F5 S14 × F16,7 S5 × F3 S14 × F24,7 S7 × F2 S8 × F2 S9 × F2 S38× F45,1915 F32,15 S3 × F5 S4 × F4 S5 × F3 S15 × F12,5 F112,15 S8 × F2 S9 × F2 S15 × F8,316 S2 × F8 F51,16 S4 × F4 F85,16 S6 × F3 F126,55 S8 × F2 S9 × F2 S10 × F217 F36,17 S3 × F6 F72,17 F95,39 S6 × F3 F140,41 F144,17 S9 × F2 S10 × F218 S2 × F9 S3 × F6 S18 × F20,9 S5 × F4 S6 × F3 S7 × F3 S18 × F40,9 S9 × F2 S10 × F219 F40,19 F60,19 S4 × F5 S5 × F4 F120,19 S7 × F3 F176,65 F180,19 S10 × F220 S2 × F10 S3 × F7 S4 × F5 S5 × F4 S20 × F33,10 S7 × F3 S20 × F12,5 F198,89 S10 × F2

11 12 13 14 15 16 17 18 19 20

2 F24,11 S2 × F6 F28,13 S2 × F7 F32,15 S2 × F8 F36,17 S2 × F9 F40,19 S2 × F103 S3 × F4 S3 × F4 F42,13 S3 × F5 S3 × F5 F51,16 S3 × F6 S3 × F6 F60,19 S3 × F74 S4 × F3 S4 × F3 F56,13 S14 × F16,7 S4 × F4 S4 × F4 F72,17 S18 × F20,9 S4 × F5 S4 × F55 F60,11 F70,29 S5 × F3 S5 × F3 S5 × F3 F85,16 F95,39 S5 × F4 S5 × F4 S5 × F46 S6 × F2 S6 × F2 F84,13 S14 × F24,7 S15 × F12,5 S6 × F3 S6 × F3 S6 × F3 F120,19 S20 × F33,107 S7 × F2 S7 × F2 S7 × F2 S7 × F2 F112,15 F126,55 F140,41 S7 × F3 S7 × F3 S7 × F38 F112,41 S12 × F8,3 S8 × F2 S8 × F2 S8 × F2 S8 × F2 F144,17 S18 × F40,9 F176,65 S20 × F12,59 F117,53 S12 × F15,4 F153,35 S9 × F2 S9 × F2 S9 × F2 S9 × F2 S9 × F2 F180,19 F198,89

10 F120,11 S12 × F35,6 F160,49 S38× F45,19 S15 × F8,3 S10 × F2 S10 × F2 S10 × F2 S10 × F2 S10 × F211 S11 F143,12 F165,76 F187,67 F209,56 F231,43 S11 × F2 S11 × F2 S11 × F2 S11 × F212 F143,12 S12 F168,13 S14 × F48,7 S15 × F24,5 S16 × F15,4 F264,109 S18 × F8,3 S12 × F2 S12 × F213 F165,76 F168,13 S13 F195,14 F221,103 F247,77 F273,64 F299,116 F325,51 S13 × F214 F187,67 S14 × F48,7 F195,14 S14 F224,15 S16 × F63,8 F280,99 S68× F77,34 F336,71 S54× F91,2715 F209,56 S15 × F24,5 F221,103 F224,15 S15 F255,16 F285,134 S18 × F35,6 F345,91 S20 × F15,416 F231,43 S16 × F15,4 F247,77 S16 × F63,8 F255,16 S16 F288,17 S18 × F80,9 F352,111 S20 × F24,517 S11 × F2 F264,109 F273,64 F280,99 F285,134 F288,17 S17 F323,18 F357,169 F391,13718 S11 × F2 S18 × F8,3 F299,116 S68× F77,34 S18 × F35,6 S18 × F80,9 F323,18 S18 F360,19 S20 × F99,1019 S11 × F2 S12 × F2 F325,51 F336,71 F345,91 F352,111 F357,169 F360,19 S19 F399,2020 S11 × F2 S12 × F2 S13 × F2 S54× F91,27 S20 × F15,4 S20 × F24,5 F391,137 S20 × F99,10 F399,20 S20

Table 19: Colorings that realize thenumber of colors in Table 18.

various topics 137

Proof.If. If m divides n, then Sm × F1,n is a sufficient coloring with mn

colors, and therefor it is a sufficient coloring, and so C(m, n) = mn.Only if. Assume then m < n, and that we have a sufficient coloring

using mn colors. Consider a (m + 1)× (n + 1) rectangle.Let’s say the color in the top-left corner is purple. All the colors

in the leftmost n columns of the top row will be called “shades ofred”, and all the colors in the top m columns of the left column willbe called “shades of blue”, as shown in Figure 139. Purple is both ashade of red and a shade of blue.

Figure 139: Proof

We now look at row m + 1. The only colors available for the left-most n columns are shades of red. Furthermore, as m < n, the left-most square in row m + 1 cannot be purple, as this would cause avertical rectangle with the same upper-left corner to have two purplesquares. With only the shades of red available for that row, purplemust appear somewhere else among the leftmost n columns in rowm + 1.

On the other hand, in column n + 1 we can only use shades ofblue, among which there must be a purple square. If the circledsquare does not use the color purple, then the lower-right (m × n)rectangle has two purple squares. Hence the circled square must bepurple. Thus two squares at distance n along the same row musthave the same color. Repeating the argument with rows and columnsinterchanged shows that two square at distance n along a columnhave the same color.

We have shown that if two squares lie in the same row or the samecolumn and are exactly n squares apart on that row or column, thenthey must both have the same color. Note that since no interveningsquares on that row or column can also have the same color, thismeans that every row and every column is basically colored periodi-cally, with period n.

We next prove that m divides n.Suppose that m does not divide n, but we have an mn-coloring.

This mn-coloring is determined by its values on an n × n square.Let Ci be the set of colors used on the ith row of this square. We seethat C1, . . . , Cm are pairwise disjoint (these rows all being containedwithin an m × n rectangle), and that Ci = Cm+i for all i < n − m,since Cm+i must be disjoint from Ci+1, . . . , Cm+i−1, leaving only the ncolors in Ci available. (Row m + i and row i might have these colorsin a different order, but they will be the same set of colors.)

If m divided n, then we’d get each of the sets C1, . . . , Cm appearingexactly n/m times on the square. However, since m doesn’t dividen, this repeating pattern of sets gets “cut off” at the bottom, and C1

appears on some row Cn−i for i < m. Now a horizontal rectangle

138 polyominoes

starting at row n− i will contain two rows colored using colors fromC1 once the square repeats, contradicting the hypothesis that this is asufficient coloring.

Therefor, if we have a sufficient coloring using mn colors, m mustdivide n.

Theorem 148 (Puleo (2018b)2). For m > 1, C(m, m + 1) = m(m + 2). 2 Some details of the proof have beenprovided by Peter Taylor; see thereference for more information.


Proof. Consider a m × (m + 1) rectangle. Let all the top colors beshades of red, and the right-most color crimson.

Let all the colors not in the rectangle be shades of yellow. The cellsbelow the rectangle must either be shades of yellow, or shades of red.But consider now a m × (m + 1) rectangle with its top right cornerjust to the left of the crimson cell. We can see the cells must all becrimson or a shade of yellow, except the right-most one that mustbe purple. By moving this rectangle left, we can see the only one ofthose that can be crimson is the left most.

Figure 140: Proof

Now suppose we have less than m shades of yellow. This meansneither the left-most or right-most cell cannot be yellow. They mustbe crimson and purple respectively. This means we have a periodicalcoloring, with a cell the same color as those with offset (±m,±m).

Therefore the tiling is periodic: every mth row consists of the m + 1reds and m1 yellows shifted by m from the row m before. But noneof the rows in between can contain any red or yellow, so we end uprequiring m(2m) = 2m2 colors. But if we have less than m shadesof yellow, we have less than m(m + 1) + m colors (the colors insidethe purple rectangle plus the yellows), i.e. we have less than m2 + 2mcolors. But for m > 1, this is less than 2m2, which is a contradiction.Therefor we need at least m shades of yellow, and so we need at leastm(m + 2) colors in total.

But we also have C(m, m+ 1) ≤ m(m+ 2) (Theorem 150), and sincewe need at least m(m + 2) colors, we need exactly m(m + 2) colors.

The coloring Fm(m+1)−1,m(m+2) is sufficient, and since it uses m(m+

2) colors it is also optimal.

Theorem 149. Suppose P is a scaled copy of Q, and the scaling factor is a.Then C(P) = n2C(Q). If K is the optimal coloring for Q, then K × Sa isthe optimal coloring for P.

[Not referenced]

Proof.

various topics 139

Theorem 150. Assume m ≤ n. Then C(m, n) ≤ (m′)2k where m′ ≥ mand m′k ≥ n, with Sm′ × Fk a sufficient coloring.


Proof. If m′ ≥ m and m′k ≥ n, then R(m, n) ⊂ R(m′, m′k), andso C(m, n) ≤ C(m′, m′k) (Theorem 140), and so C(m, n) = (m′)2k(Theorem 147).

Theorem 151 (Taylor (2018)). C(am, an) ≤ a2C(m, n)

[Not referenced]

Proof. Let K be an optimal coloring for R(m, n). Define K′(x, y) =

a2K(⌊ x

a⌋

,⌊ x

a⌋)+ (x mod a)+ a(y mod a). This coloring uses a2C(m, n)

colors.It is easy to see K′ is sufficient for R(am, an), and therefor C(am, an) ≤

a2C(m, n).

Theorem 152. C(k, 2k− 1) = 2k2


Proof. R(k, 2k− 1) ⊂ R(k, 2k), so C(k, 2k− 1) ≤ C(k, 2k) = 2k2. So allwe need to show is that 2k2 − 1 colors are not enough.

Towards a contradiction, suppose it is enough, and that K is anoptimal coloring. Let δxy = x + y mod 2, and define a new coloring asfollows:

K′(x, y) =

K(

x+y2 , x−y

2

)for (x + y) ≡ 0 (mod 2)

K′(x− 1, y) + k2 − 1 for (x + y) ≡ 1 (mod 2)

(a)

(b)

(c)Figure 141: An example showingthe coloring transformation used inTheorem 152. The first image show (anexample) of original coloring K. Thesecond image the new coloring for cellswith x + y (other cells are left blank).The final image shows the completenew coloring, with a darker shade of acolor i used for color i + k2 − 1.

Notice that

• K′ has twice the number of colors as K, so it has 4k2 − 2 colors.

• Two cells (x, y) and (x′, y′) can only have the same color ifx + y ≡ x′ + y′ (mod 2).

• If two cells have the same color in K′, then two correspondingcells have the same color in K.

We will now show K is an sufficient coloring for R(2k, 2k − 1).Suppose it is not; then there are two cells that can be covered withR(2k, 2k− 1). These cells must satisfy x + y ≡ x′ + y′ (mod 2), andWLG assume x + y ≡ x′ + y′ ≡ 0 (mod 2).

140 polyominoes

Now if (x, y) and (x′, y′) lies inside the same R(2k, 2k − 1), thena and b lie inside the same R(k, 2k − 1), which means K cannot besufficient, a contradiction. Therefor, K′ must be sufficient.

Now K′ uses 4k2 − 2 colors, but by Theorem 148 C(2k, 2k − 1) =

4k2 − 1; this is impossible (no sufficient coloring can use less colorsthan the chromatic number, which is minimal by definition). Thismeans then that 2k2 − 1 colors is not enough for R(k, 2k− 1), which iswhat we wanted to show.

Therefor, C(k, 2k− 1) = 2k2.

Figure 142: The first 8 W-polyominoes

Problem 42. The theorem above “scaled” the coloring above by a factorof two, so that we could deduce constraints on the optimal colorings forR(k, 2k− 1) from R(2k, 2k− 1) (rectangles double the size).

It is worth investigating why this idea cannot be modified to work inother cases:

• We can the double coloring not work to make deductions for R(2k, 2k+1)?

• Why can we not use the trick above twice, for example, to deducesomething for R(k, 4k + 1) from R(2k, 4k + 1)?

• The transformation preserved squares (in the sense that a squareof colors from the same colors will transform into a square of samecolors in the new coloring). Why is this necessary?

• There are not square-preserving transformations that are not simplyscale-transformations for all factors. What factors are possible?

• Why can other factors than 2 not work; for example, to try to deducesomething for R(k, 5k− 1) from R(5k, 5k− 1)?

• We can use a forward and backward translation to get rational fac-tors that does not have the problem of other integer factors. Why canthis idea not work?

3 Since formulating this conjecture Idiscovered that it is not completelytrue. The first part seems true (whenm <

√2n, but the last part is false in

between a third and a half of cases.

Problem 43 (Conjecture3). Prove that for m ≤ n

C(m, n) =

mn−m2 if m <√

2n

m2 ⌈ nm⌉

otherwise.

Note that we already proved it for when m|n, when n = m − 1,and when n = 2m− 1.

A general W-polyomino, which we will denote by Wn, is a snakewith n cells that alternate in two directions. 4 W-polyominoes are

4 Scherphuis (2016) discovered thatthe set of tiles W1, W2, · · ·W8 make aninteresting puzzle. The set tiles manysymmetric figures, including R(6, 6),R(4, 9), B(1 · 2 · 3 · 4 · 5 · 6 · 7 · 8),B(3 · 4 · 5 · 62 · 5 · 4 · 3), B(5 · 6 · 72 · 6 · 5).He calls these polyominoes zig-zagpolyominoes.

weak. W polyominoes with odd area are unusual because they arenot efficient.

various topics 141

Theorem 153. For Wn, and let k = dn/2e. We need 2k colors, and anoptimal coloring is given by F2k,k.


Proof. It is clear that F2k,k is has enough colors. Since it has 2k colors,and 2k = n for n even, it follows that C(P) = n for n even. It remainsto show that for odd n, when 2k = n + 1, we need n + 1 colors.

Since Wn ⊂ Wn+1, it follows that C(Wn) ≤ C(Wn+1) (Theorem140), so for odd n, C(Wn) ≤ C(Wn+1) = n + 1. It is therefor enoughto show that n colors are not enough.

Let’s assume we need only n colors. Place the W-polyomino any-where, and color the cells that it covers from 0 to n− 1.

Figure 143: Coloring the first cells.

Now place the W-polyomino so that it covers all the same cells ex-cept the one with color 0. The uncolored cell must be of color 0 (sincewe have only n colors and all the other colors are used already).

Figure 144: One color is forced.

Repeat with color 1.

Figure 145: Another color is forced.

We can continue in this way, but for this proof we need to go onlythis far.

Now put the W-polyomino so that it covers the first few cells witheven color, but none with odd colors. The cells that are not coloredyet must be colored with the odd colors, one of each (again, becausewe only have n colors). In particular, one of the cells must be color1. It cannot be the cell next to color 0, since it lies in the same Wshape as the cell below which also has color 1. Therefor, one of theremaining cells must be color 1. Those cells are marked orange inFigure 146.

Figure 146: One of the unnumberedorange cells must be color 1.

Finally, move the W-polyomino one cell down and one to the right.It now covers all three orange cells (one of which must be color 1),and another cell with color 1. This violates the condition of this be-ing a legal coloring, which means, we cannot get away with only ncolors.

Figure 147: Two of the same colors(color 1) must occur in the same W.Therefor, we need at least one colormore.

Problem 44. Show that

(1) the only efficient pentominoes are the I- and X-pentominoes;

(2) the only efficient hexominoes are the I- and W-hexominoes; and

(3) the only efficient heptominoes is the I-heptomino.

While efficient polyominoes are relatively scarce for polyominoeswith 7 or less cells, there are lots with 8 or 9 colors; the problem be-low explores why.

Problem 45.

142 polyominoes

(a) Checkerboard coloring (F2) (b) S2 (c) F3

(d) F6,3 (e) F5,2 (f) S2 × F2

(g) S3 (h) F8,5 Figure 148: Some standard colorings.

various topics 143

(1) Find a coloring Fn,k that is not the optimal coloring of any efficientpolyomino.

(2) Show that any polyomino formed from k2 cells of different colorsfrom a square coloring Sk has that square coloring as an optimalcoloring, and hence such an polyomino is efficient.

(3) Show that any polyomino formed from 2k2 cells of different colorsfrom a coloring F2 × Sk has that coloring as an optimal coloring, andhence such an polyomino is efficient.

(4) Find examples that show the above in general does not hold for poly-ominoes with mk2 cells of different colors from Fm,n × Sk.

(5) How many octominoes are efficient?

(6) Can you say which decominoes are efficient?

Problem 46.

(1) Is every optimal coloring periodic?

(2) Does every polyomino have an optimal coloring of the form Fm,n ×Sk?

(3) We considered colorings where each polyomino cover each color atmost once. What about colorings where each polyomino covers eachcolor at most k times? At least k times?

(4) Complete the tables by finding the exact chromatic number for theunknown hexominoes and heptominoes.

144 polyominoes

Polyomino P C(P) Theorems

Monomino 1 F1 147(1)

Domino 2 F2 147(1)

Bar 3 F3 147(1)Right 4 S2 141

Bar 4 F4 147(1)Square 4 S2 147(2)

T 5 F5,2 139 (Q = X5), 143 (Q = X5)L 8 S2 × F2 141 (Q = R(3, 2)), 137

Skew 4 F4,2 153

I 5 F5 147(1)L 8 S2 × F2 141

P 8 S2 × F2 140 (Q = L4), 137

Y 8 S2 × F2 140 (Q = L4), 137

N 8 S2 × F2 140 (Q = L4), 137

W 6 F6,3 153

F 8 S2 × F2 140 (Q = L4), ColoringV 9 S3 141

T 8 S2 × F2 140 (Q = L4), ColoringU 8 S2 × F2 140 (Q = L4), 137

X 5 F5,2 144

Z 9 S2 × F2 139 (Q = R(3, 3)), 137

[8, 12]8 S2 × F2 139 (Q = L4), 137

8 S2 × F2 139 (Q = L4), 137

8 S2 × F2 139 (Q = L4), 137

(8) S2 × F2 140 (Q = T5)9 S3 140 (Q = V5), 137

8 S2 × F2 148

12 S2 × F3 141

9 S3 140 (Q = V5), 137

6 F6 147(1)

Table 20: Chromatic numbers for smallpolyominoes. Intervals [a, b] denotelower and upper bounds. Numbersin brackets show the upper bound isprovided by an explicit coloring ratherthan a theorem.

various topics 145

Polyomino P C(P)

9 S3 140 (Q = Z5), 137

8 S2 × F2 140 (Q = L4), 137

8 S2 × F2 140 (Q = L4), Coloring(9) S3 139 (Q = V5), Coloring6 F6,3 153

(9) S3 140 (Q = V5), Coloring

(9) S3 140 (Q = V5), Coloring(9) S3 140 (Q = Z5), Coloring

15 F15,11 141

(8) S2 × F2 140 (Q = L4), Coloring(8) S2 × F2 140 (Q = L4), Coloring9 S3 140 (Q = Z5), 137

(8) F8,5 140 (Q = L4), Coloring(8) S2 × F2 140 (Q = L4), Coloring

8 S2 × F2 140 (Q = L4), 137

[9, 15](8) S2 × F2 140 (Q = L4), Coloring(8) S2 × F2 140 (Q = L4), Coloring[9, 10] (F10,7)(9) S3 139 (Q = V5), Coloring(8) S2 × F2 140 (Q = L4), Coloring

(8) S2 × F2 140 (Q = L4), Coloring

(8) S2 × F2 140 (Q = L4), Coloring8 S2 × F2 140 (Q = L4), Coloring

(8) S2 × F2 140 (Q = L4), Coloring

Table 21: Chromatic numbers for smallpolyominoes.

146 polyominoes

Polyomino P C(P)

7 F7 147(1)12 S2 × F3 141

12 S2 × F3 140 (Q = L6), 137

12 S2 × F3 140 (Q = L6), 137

18 S3 × F2 141

[8, 12]12 S2 × F3 140 (Q = L6), 137

(8) S2 × F2 140 (Q = L4), Coloring

[8, 18]

[15, 18]

(15) F15,4 140 (Q = L(3, 4)), Coloring

(15) F15,4 140 (Q = L(3, 4)), Coloring

16 S4 141

12 S2 × F3 140 (Q = L(2, 5)), 137

[8, 12](8) S2 × F2 140 (Q = L(2, 5)), Coloring

15 F15,4 140 (Q = L(3, 4)), 137

[8, 12]

[8, 18]

[8, 18]

[8, 18]

[8, 16]

[9, 18]

[9, 15]

[9, 15]


various topics 147

Polyomino Class

(9) S3 140 (Q = V5), Coloring

(9) S3 140 (Q = V5), Coloring

(9) S3 140 (Q = V5), Coloring

[9, 16]

[15, 16]

(15) F15,4 140 (Q = L(3, 4)), Coloring

(15) F15,4 140 (Q = L(3, 4)), Coloring

(15) F15,4 140 (Q = L(3, 4)), Coloring

15 F15,4 140 (Q = L(3, 4)), 137

12 S2 × F3 140 (Q = L(2, 5)), 137

[8, 12](8) 140 (Q = L4), Coloring

15 F15,4 140 (Q = L(3, 4)), 137

8 S2 × F2 140 (Q = L4), 137

(9) S3 140 (Q = V5), Coloring

[9, 15]

(9) S3 140 (Q = V5), Coloring

(9) S3 140 (Q = V5), Coloring[8, 12]

[8, 15][8, 12]

[8, 18]

[8, 18]

(9) S3 140 (Q = V5), Coloring

[8, 15]


148 polyominoes

Polyomino P C(P)

[8, 18]

(9) S3 140 (Q = V5), Coloring

8 F8,4 153

[8, 16]

[8, 16]

[9, 16]

[9, 15], 12?

[8, 15]

(9) S3 140 (Q = V5), 137

[9, 15]

(9) S3 140 (Q = V5), Coloring

[9, 15]

[8, 18]

(9) S3 140 (Q = V5), Coloring

(9) S3 140 (Q = V5), Coloring

(9) S3 140 (Q = V5), Coloring

[9, 16], 15?

(9) S3 140 (Q = V5), Coloring

(8) S2 × F2 140 (Q = L4), Coloring

(9, 15)

(8) S2 × F2 140 (Q = L4), Coloring

[9, 16]

[9, 15]

(8) S2 × F2 140 (Q = L4), Coloring

[11?, 16]


various topics 149

Polyomino Class

[9, 15]

[12, 18]

[12, 18]

[12, 18]

[12, 18]

[12, 18]

[8, 15]

[8, 9]

[9, 15]

15 F15,4 140 (Q = L(3, 4)), 137

12 S2 × F3 140 (Q = L(2, 5)), 137

8 S2 × F2 140 (Q = L4), 137

[9, 15]

[8, 15]

8 S2 × F2 140 (Q = L4), Coloring

9 S3 140 (Q = Z5), 137

(8) S2 × F2 140 (Q = L4), Coloring

9 S3 140 (Q = V5), 137

9 S3 140 (Q = V5), 137

[8, 15]

[9, 15]

(8) S2 × F2 140 (Q = L4), Coloring

8 S2 × F2

9 S3 140 (Q = V5), 137

9 S3 140 (Q = V5), 137


150 polyominoes

Polyomino P C(P)

(8) S2 × F2 140 (Q = L4), Coloring

[9, 18]

[8, 18]

[9, 16]

[9, 16]

[8, 18]

[9, 16]

[8, 18]


various topics 151

5.3 Border Words

Simply-connected polyominoes can be uniquely defined by their bor-der; and the border can be described as a set of unit steps in one offour directions. Let’s use x for horizontal steps, and y for verticalsteps. And let’s use an exponent for the amount and direction; posi-tive for right and up, negative for left and down. Finally, let’s decideto always go anticlockwise around the polyomino with the inside onthe left. We call this sequence of steps a word.

With this, one way to describe a monomino is xyx−1y−1. We canalso start with another vertex, giving us a cyclically shifted wordsuch as yx−1y−1x. We will not distinguish between words cyclicallyshifted. Fig 27 shows the border words for the tetrominoes.

Polyomino Border word

x2y2x−2y−2

xyxyx−3y−1xy−1

x2yx−1yx−2y−1xy−1

x3y2x−1y−1x−2y−1

x4yx−4y−1

Table 27: Border words for tetrominoes.

Problem 47.

(1) What is the border word of R(m, n)?

(2) What is the border word for B(ma · nb)?

(3) Prove sum of exponents of x equals the sum of exponents of y equals0.

(4) Suppose we have a word to describe a segment xayb; what is theword that describes the same segment traveled in reverse?

The reverse of a word W = xa1 yb1 xa2 yb2 · · · xan ybn is defined asW = y−bn x−an · · · y−b2 x−a2 y−b1 x−a1 ; that is, we reverse the order ofthe steps and change the direction of all steps.

If W is the border word of a polyomino, then W is the word of aninfinite region with a hole the same shape as the original polyomino.

Problem 48. Suppose W = xa1 yb1 xa2 yb2 · · · xan ybn is a polyomino. Whathappens if we

152 polyominoes

(1) scale all exponents of x by a factor k?

(2) scale all exponents by a factor k?

(3) change the signs of all exponents of x?

(4) change the signs of all exponents of x and y?

(5) swap x and y?

Can you write conditions on the border word of a polyomino for whichtype of symmetry the polyomino has?

6Rectangles

We saw that determining whether a region can be tiled by dominoescan be determined efficiently. But in general,this is not the case. Infact, it is hard to determine if a polyomino will even tile a rectangle.

Consider the right tromino. It is easy to see it will tile R(2, 3),and thus all rectangles R(2m, 3n). Are these the only rectanglesthat it will tile? It turns out we can also tile R(5, 9), so and so alsoR(5m, 9n). Have we found all the rectangles it will tile? Not yet! Forwe can stack R(2, 6) and R(3, 6) together to tile R(5, 6), which addsR(5m, 6n) to the mix. Combined with the R(5m, 9n), this allows us totile R(5m, 3n) for n ≥ 2. And combining this in turn with R(2m, 3n)allows us to tile R(m, 3n) for m = 2 and all m > 3 and n > 2. Thus,we can tile almost all rectangles with area divisible by three (the onlyexceptions are R(m, 3) for odd m — which are in fact not tileable aswe will see later).

The tromino case was easy enough to analyze. We found tworectangles, called prime rectangles, from which all other tilings can bederived. But in other cases, the analysis is much more difficult. Forexample, the Y-pentomino requires a set of 40 rectangles before wecan construct the set of all rectangles.

And what is really surprising, is how difficult it is to know whichrectangles can be tiled by a set of rectangles. This is the topic of thefirst section.

The minimum number of polyominoes that can tile a rectangle iscalled the order of the polyomino. We know very little about whatorders are possible. We know orders 2 and 4 are possible (and in factubiquitous), and we know order 3 is impossible. But we do not knowwhether 5, or 6, or 7 are possible. We know all orders of the form4k are possible, but other than that, the highest order polyomino weknow is 246. This is the topic of the second section.

The third section deals with prime rectangles and proofs of poly-ominoes that don’t tile rectangles.

The fourth section deals with fault-free tilings of polyominoes.

154 polyominoes

While we have a nice solution for when the polyomino is a rectangle,in general the picture is murky. The easy cases are usually those witha small number of prime rectangles.

The final section deals with simple tilings — tilings that haveno subtilings with more than one tile that are rectangles. There aresimple tilings of rectangles by any number of pieces larger than for,except for 6. (This is the second time 6 shows up out of the blue as anexception — recall that R(m, n) has fault free tilings form m, n > 4,except for R(6, 6). This is probably a coincidence.)

6.1 Tilings by Rectangles

6.1.1 Which rectangles can be tiled by a set of rectangles?

Many problems can be broken into simpler problems. One naturalway to approach a polyomino tiling problem is to break regionsinto rectangles. Rectangles are the simplest figures, and thereforwe can say a lot about them. This section is primarily about howrectangles can tile other rectangles. (So, if we know that a polyominotiles certain rectangles, we can answer which other rectangles it maytile.) Unfortunately, this is a rather hard problem to solve in general.

In this section throughout T = {R(pi, qi)} is a set of rectangles,and R = R(m, n) is some rectangle we wish to tile using tiles fromthe set without rotation. If a tiling problem allows for rotation, simplyadd a rotated copy to the set of rectangles. We will prove a bunchof theorems; Table 28 at the end of the section summarizes whichtheorem applies in each situation.

Figure 149: The smallest rectangle thatrequires all three tiles from the setR(2, 3), R(3, 2) and R(5, 5).

Figure 150: The smallest “interesting”rectangle that requires all three tilesfrom the set R(2, 3), R(3, 2) and R(5, 5).

We begin by stating obvious necessary conditions. Let T =

{R(pi, qi)}. Then T can tile R(m, n) when all the following condi-tions are met:

rectangles 155

(1) m = ∑ piai for some integers ai ≥ 0 (Theorem 17)

(2) n = ∑ qibi for some integers bi ≥ 0 (Theorem 17)

(3) mn = ∑ piqici for some integers ci ≥ 0 (Theorem 1)

Example 16.

• If T = {R(2, 3), R(3, 2)}, and R = R(5, 5), then conditions (1) and(2) hold, but not (3).

• If T = {R(2, 3), R(3, 3)}, and R = R(3, 5), then conditions (1) and(3) hold, but not (2).

The following theorem is trivial; I state it for completeness.

Theorem 154 (Bower and Michael (2004), Observation, Section 1). LetT = {R(p, q)}. A tiling exists if and only if p | m and q | n.

[Referenced on pages 157, 158, 161, 201 and 251]

Proof.If. Suppose m = pk and n = q`. A tiling is given by placing ` rows

with k tiles in each row.Only if. All sides of the rectangle are mountains, so necessity fol-

lows directly from Theorem 17.

Theorem 155 (Klarner (1969), Theorem 5). Let T = {R(p, 1), R(1, p)}.A tiling exists if and only if p divides m or n.


Proof.If. Suppose WLG that p divides m, so that m = pk. We can put k

bars in a row, and stack n of them to form a tiling of R(m, n).Only if. Apply the flag coloring with p colors (that is, each cell

with coordinates x, y gets the color (x + y) mod p).No matter how we place the bars, each covers one each of the

p colors. So if the tiling requires k tiles, a tiling will have k of eachcolor.

Now suppose m mod p = m′ > 0 and n mod p = n′ > 0, andWLG that m′ ≤ n′. We can divide the rectangle into 4 quadrants:R(m − m′, n − n′), R(m − m′, n′), R(m′, n − n′) and R(m′, n′). Thefirst three have tilings by the bars (by the first part of the theorem)and so each have the same number of each color. The last rectangle,however, has m′ of color 0, but m′ − 1 of color 1 (see Figure 151).Therefor, a tiling is impossible, and therefor either m′ or n′ must be 0,and so either m or n must be divisible by p.

Figure 151: An example with m′ = 5,n′ = 7, and p = 8. There are 5 cellsof light blue, but only 4 cells of green-blue.

156 polyominoes

Theorem 156 (Klarner (1969), Corollary of Theorem 5). Let T =

{R(p, q), R(q, p)}. A tiling exists if and only if each of p and q divideeither m or n, and if pq divides one of side, the other side can be expressed aspx + qy with x, y ≥ 0.


Proof.If. To prove there is a tiling when the conditions hold, there are

two cases to consider:

(1) p|m and q|n

(2) p|m, q|m and n = px + qy.

Suppose p < q. The first case is trivial: simply place m/p rectanglesin n/q rows. For the second case: a tiling is given by placing x rowsof m/p rectangles vertically, and y rows of m/q rectangles horizon-tally.

Only if. If R(m, n) has a tiling by T , then it must also have a tilingby R(p, 1) and R(1, p), and so by Theorem 155 we must have p di-vides m or n.

Similarly, it must also have a tiling by R(q, 1) and R(1, q), and so qdivides m or n.

Each side is expressible in the form px + qy by either Theorem 17

or 18.

Theorem 157 (Slightly generalized from Martin (1991), p.43 ). R(m, n)is tileable by R(p, 1) and R(1, q) iff p divides m or q divides n.


Proof. If. WLG suppose p divides m. A tiling is given by stackingR(p, 1) in each row.

Only if. WLG, assume p does not divide m. We will show q dividesn.

Color the columns of R(m, n) cyclically with p colors. Denote thenumber cells of color i by ci. Since p does not divide m, we can writem = dp + r, where 0 < r < p. There is d + 1 columns of the firstcolor, and d columns of the second color; therefor c0 = n(d + 1), andcp−1 = nd.

Consider now a tiling by the two bars. The horizontal bar coversone of each color; the vertical bar covers q cells of the same color. Letx be the number of horizontal bars, and yi the number of vertical barsthat cover cells of color i.

rectangles 157

Then c0 = x + qy + 0, and cp−1 = x + qyp−1, so c0 − cp−1 =

q(y0 − yp−1).Now

n = n(d + 1)− hd

= c0 − cp−1

= q(y0 − yp−1),

which means q divides n, which is what we wanted to show.

Theorem 158 (Divisibility Lemma, Bower and Michael (2004), Section3). Suppose we have a set of rectangles partitioned into two sets T1 and T2

such that all rectangles in T1 have widths with a factor r and all rectanglesin T2 have heights with a factor s. Then if T tiles a R(m, n), then r | m ors | n.


Proof. 1 The tiles in T1 are all tileable by R(r, 1), and the tiles in T2 1 The original proof uses the idea ofcharges. The argument here hingeson Theorem 157, where we adaptedan argument Martin originally usedto prove Theorem 155 (and is in factvery similar to the argument we usedthere from Golomb). Clearly these threetheorems are closely related.

are all tileable by R(1, s). Therefor, if a tiling of R(m, n) by T exists, italso has a tiling by T ′ = {R(r, 1), R(1, s)}.

But by Theorem 157, this is possible only if r | m or s | n.

Theorem 159 (de Bruijn (1969), Theorem 1). Suppose a number k dividesone side of each rectangle in T . If T tiles R(m, n), then either k | m ork | n.


Proof. This follows immediately from 158 by setting r = s = k.

One consequence of this theorem is that we cannot tile a 10× 10with horizontal and vertical bars of length 4.

Theorem 160 (Bower and Michael (2004), Theorem 5). Let p1, p2, q1,and q2 be positive integers with gcd(p1, p2) = gcd(q1, q2) = 1. ThenR(m, n) can be tiled by translates of R1 = R(p1, q1) and R2 = R(p2, q2)

rectangular bricks if and only if one of the following holds:

(1) p1 | m and q1 | n

(2) p2 | m and q2 | n

(3) p1 p2 | m and n = aq1 + bq2 for some integers a, b

(4) q1q2 | n and m = ap1 + bp2 for some integers a, b.


158 polyominoes

Proof. 2 2 Kolountzakis (2004) gives an alterna-tive proof using the Fourier transform.Fricke (1995) showed the special casewhen the two rectangles are squares.

If. In case (1), a tiling of R(m, n) by R(p1, q1) exists by Theorem154; similarly in case (2), a tiling ofR(m, n) by R(p2, q2) exists.

In case (3), suppose m = m′p1 p2. We can split the rectangle inR(m′p1 p2, aq1) and R(m′p1 p2, bq2). From Theorem 154, the first istileable by R1 and the second by R2.

In case (4), suppose n = n′q1q2. We split the rectangle in R(ap1, nq1q2)

and R(bp2, nq1q2), which by Theorem 154 is tileable by R1 and R2 re-spectively.

Only if. If R is tileable by R(p1, q1) and R(p2, q2), then m = ap1 +

bp2 for some integers a, b and n = a′q1 + b′q2 for some integersa′, b′ (Theorem 17 or 18). Since R1 has width p1, and R2 has heightq2, by the divisibility lemma (Theorem 158) we have p1 divides m orq2 divides n. Similarly, R2 has width p2 and R1 has height q1, so p2

divides m or q1 divides n. Taken together, one of the four conditionsmust hold.

Problem 49. Which rectangles can be tiled by R(2, 2) and R(3, 3)?

Example 17. Let T = {R(6, 6), R(10, 10), R(15, 15)}. There are 6different ways to partition this set so that the conditions for Theorem 158hold. These give the following set of conditions, which must be satisfied forany tileable rectangle R(m, n):

(1) 2 | m or 15 | n

(2) 3 | m or 10 | n

(3) 5 | m or 6 | n

(4) 15 | m or 2 | n

(5) 10 | m or 3 | n

(6) 6 | m or 5 | n

From this we can make inferences such as the following:

• If m = 6, it already satisfies 1, 2, and 6. Since none of 5, 15, or 10divides 6, the only way to satisfy conditions 4, 5 and 6 is for 6 | n.

• For a tileable rectangle R(m, n), we have one of 6, 10 or 15 mustdivide either m or n.

• If m is prime, then 30 | n.

It is not hard to see the conditions are sufficient for a tiling to exist providedthe sides satisfy Theorem 17.

rectangles 159

Example 18. Let T = {R(3, 2), R(2, 3), R(9, 5), R(5, 9)}. These rectangleshave a common factor 3, so we know they will only tile rectangles with afactor of 3.

We can stack three of the smaller rectangles to make R(9, 2) and R(2, 9).By combining these with R(9, 5), we R(5, 9), we can tile R(9, k) or R(k, 9)for any k 6= 1, 3.

We can also tile R(6, 2), R(2, 6), and we can also tile R(6, 3) andR(3, 6), so we can tile any R(6, k) and R(k, 6) for any k > 1. (Becauseany k > can be written as 2x + 3y for nonnegative integers x and y.)

Finally, we can tile all R(3, 2k) and R(2k, 3) for k ≥ 1. Putting thistogether, we have the following:

(1) R(3m, n) for all m > 1, n > 3

(2) R(3, 2n) for all n ≥ 1

(3) R(2m, 3) for all m ≥ 1

Theorem 161 (Fricke (1995)). Let a,b, x and y be positive integers withgcd(x, y) = 1. Then, R(a, b) can be tiled with R(x, x) and R(y, y) if andonly if either

(1) a and b are both multiple of x, or

(2) a and b are both multiple of y, or

(3) one of the numbers a, b is a multiple of xy and the other can beexpressed as a nonnegative integer combination of x and y.

[Not referenced]

Things get more interesting when T has more than two rectan-gles. For rectangles satisfying certain conditions, they can tile anysufficiently large rectangle.

This is somewhat similar to what happens to integers. Given twointegers that are relatively prime3, we can write any big-enough 3 Recall, two integers are realtively

prime when they don’t have any com-mon divisors other than 1

integer as a positive combination of the two. For example, givenintegers 3 and 5, we can write any integer larger than 7 as the sumof multiples of 3 and 5. For example, 8 = 3 + 5, 13 = 3 + 2 · 5,11 = 2 · 2 + 5.

Of course, we can do the same when we have a larger set of inte-gers. For example, given 3, 4 and 5, we can write any number largerthan 3 as a positive combination. We call the minimum number thatcannot be written as a positive combination of the set the set’s Frobe-nius number, and we write it g(a, b, c, · · · ). For example, g(3, 5) = 7,and g(3, 4, 5) = 2.

Theorem 162 (Labrousse and Ramírez Alfonsín (2010), Theorem 5).Suppose we have k ≥ 3 rectangles Ri, where is Ri is a pi × qi rectangle, and

160 polyominoes

(1) gcd(pi, pj) = 1, for any i 6= j,

(2) gcd(qi, qj) = 1, for any i 6= j,

Let g1 be the maximum Frobenius number of pi taken (k− 1) at a time.Let g2 be the maximum Frobenius number of products q1···qk

qiqjfor some j and

i 6= j.Then if we have pi, qi ≥ max(g1, g2), the rectangle is tileable by Ri.


The proof is quite technical, and I omit it. For details, see the ref-erence. There is also a courser version of the theorom above, that ismuch easier to calculate:

Theorem 163 (Labrousse and Ramírez Alfonsín (2010), Corollary 1).Suppose we have k ≥ 3 rectangles Ri, where is Ri is a pi × qi rectangle, and

(1) gcd(pi, pj) = 1, for any i 6= j,

(2) gcd(qi, qj) = 1, for any i 6= j,

If r = maxi(pi, qi), then the rectangle is tileable by this set if m, n > r4.

[Not referenced] Rectangles Minimum Length

2× 3, 3× 2, 5× 5 30

Theorem 164. For a set of three rectangles, one of the following holds:

(1) gcd(qi, pj) = gcd(qi, qj) = 1

(2) We can partition the rectangles as in Theorem 159 (possibly, with thetwo factors equal).

[Not referenced]

Theorem 165. Let R1, · · · , R4 be four rectangles that satisfy the followingconditions:

(1) gcd(p1, p2) = r > 1

(2) gcd(p3, p4) = s > 1

(3) gcd(qi, qj) = 1, for i, j = 1, 2, 3, 4, i 6= j

(4) gcd(r, s) = 1

The there exist C such that these rectangles can tile any R(m, n) ifm, n > C.


rectangles 161

Proof. Let u = lcm(p1, p2) and v = lcm(p3, p4). We can then buildthese rectangles:

(1) u× q1

(2) u× q2

(3) v× q3

(4) v× q4

Using the first two of these, we can build any u× x rectangle for largeenough x (say x > C1) and using the last two rectangles, we can buildany v× y rectangle for large enough y (say y > C2). Since gcd(u, v) =1, we can, when x = y, build any z× x rectangle for large enough z(say z > C3, and of course we already have x, y > max(C1, C2). So wecan tile any z× x rectangle with z, x > max(C1, C2, C3).

And of course,

• C1 = g(q1, q2) = q1q2 − q1 − q2

• C2 = g(q3, q4) = q3q4 − q3 − q4

• C3 = g(u, v) = uv− u− v = p1 p2 p3 p4−p1 p2s−p3 p4rrs

and so

C = max{

q1q2 − q1 − q2, q3q4 − q3 − q4,p1 p2 p3 p4 − p1 p2s− p3 p4r

rs

}

Theorem 166. For a set of four or more rectangles, one of the followingholds:

(1) gcd(pi, pj) = gcd(qi, qj) = 1 for i 6= j.

(2) We can partition the rectangles as in Theorem 159 (possibly, with thetwo factors equal).

(3) We can select 4 rectangles that can tile a sufficiently large rectangle.

[Not referenced]

It can be tricky to keep track of the theorems and know which oneapplies in a given situation. Table 28 summarizes to conditions andthe theorems that apply.

We conclude this section with two special cases where the rectan-gles in the tileset are all squares.

162 polyominoes

|T | Conditions Theorem

1 154

2 gcd(p1, p2) = 1 160

gcd(q1, q2) = 1

2 gcd(p1, p2) = k > 1 159

≥ 3 gcd(pi, pj) = 1, i 6= j 162

gcd(qi, qj) = 1, i 6= j

3 gcd(p1, p2) = k > 1 158

gcd(pi, p3) = 1, i 6= 3

4 gcd(p1, p2) = r > 1 165

gcd(p3, p4) = s > 1gcd(qi, qj) = 1, i 6= jgcd(r, s) = 1

≥ 4 gcd(p1, p2 · · · , pk) = r 158

gcd(qk, pk+1 · · · p|T |) = s

≥ 4 gcd(pi, pj) = 1, i 6= j, i, j = 1, 2, 3, 4 162

gcd(qi, qj) = 1, i 6= j, i, j = 1, 2, 3, 4

Table 28: Characterization of tilings ofrectangles by rectangles.

Theorem 167 (Labrousse and Ramírez Alfonsín (2010), Theorem 6).All sufficiently large rectangles can be tiled with three squares whose side-lengths are pairwise relatively prime. R(a, a) can be tiled with R(a1, a1),R(a2, a2), and R(a3, a3) when

a >

(2−

3

∑1

1ai

)3

∏1

ai

or

a > 2a1a2a3 − a1a2 − a1a3 − a2a3

[Not referenced]

For example, for squares with length 2, 3, and 5, this theoremgives us a > 2.2.3.5− 2.3− 3.5− 2.5 = 60− 6− 15− 10 = 29.

Theorem 168 (Labrousse and Ramírez Alfonsín (2010), Theorem 7).Let p > 4 be relatively prime to 2, and 3. Then R(a, a) can be tiled withR(2, 2), R(3, 3) and R(p, p) if a ≥ 3p + 2.

[Not referenced]

rectangles 163

6.1.2 The gap number of rectangular tilings

Recall that the gap number of a region R is the minimum numberof monominoes of a tiling by T +, which is the tileset T plus amonomino. One would think that determining the gap number fora set of rectangles and a rectangular region is easy; however, it is notthe case. We will give some partial results in this section.

Theorem 169. Suppose R(m, n) with m, n < p and m + n < p is arectangle with the flag coloring Fp applied. Then one color does not occur atall.


Proof. The first row goes from color 0 to color m− 1; the second rowfrom color p− 1 to color m− 2; and so on, until the last row, whichgoes from p− n + 1 to m− n.

In row i, we have the first n− i colors and the last i colors; since iranges from 0 to n− 1, all colors satisfy 0 ≤ c ≤ m− 1, or 0 ≤ c < mand p− (n− 1) + 1 ≤ c ≤ p− 1, or p− n ≤ c ≤ p− 1. From the lastinequality and m + n < p, we get m < c ≤ p− 1. So all color used aresmaller than m or bigger than m; not are equal to m. So color m is notused.

Theorem 170. Suppose R(m, n) with m mod p + n mod p < p. ThenGp(m, n) ≥ (m−m mod p)(n− n mod p).

[Not referenced]

Proof. Partition the rectangle in four rectangles R1, R2, R3, R4 suchthat R4 = R(m mod p, n mod p). There is a color that does not oc-cur in R4. Each color occurs the same number of times in each ofR1, R2, R3; each color occurs (m−m′)(n− n′)/p in R1, m′(n− n′)/pin R2, and n′(n − n′)/p in R3. One color is missing in R4 (Theorem169), so that color occurs this many times in total:

(m−m′)(n− n′)/p + m′(n− n′)/p + n′(m−m′)/p

=(mn−m′n−mn′ + m′n′ + m′n−m′n′ + mn′ −m′n′)/p

=(mn + m′n′)/p.

(6.1)

So we can fit at most that many bars, which means G(m, n) ≥ mn−p(mn−m′n′)/p, or G(m, n) ≥ m′n′.

Theorem 171 (Barnes (1979), Lemma 1). Let T = {R(1, p), R(p, 1)}.Then

164 polyominoes

GT (R(m, n)) =

(m mod p)(n mod p) if m < p or n < p or m mod p + n mod p < p

(m−m mod p)(n− n mod p) otherwise


Theorem 172. Exactly⌊

mp

⌋ ⌊np

⌋squares R(p, p) can fit in R(m, n).


Proof. (Adapted from JimmyK4542 (2017).) Apply the square color-ing Sp to R(m, n). No matter how we place a p× p square, it alwayscovers p of each color. Let c be the color at coordinates (p− 1, p− 1).This color occurs

⌊mp

⌋ ⌊np

⌋times, so we can fit at most this number of

squares.

Theorem 173. Let T = {R(p, p)}. Then

GT (R(m, n)) = mn− (m−m mod p)(n− n mod p).


Proof. By Theorem 172 we can fit at most⌊

mp

⌋ ⌊np

⌋squares, which

means the gap number satisfies G ≥ GT (R(m, n)) = mn − (m −m mod p)(n− n mod p). The naive tiling has a gap of this size, andso is optimal, and therefor the gap number is given by GT (R(m, n)) =mn− (m−m mod p)(n− n mod p).

Theorem 174.⌊

mp

⌋ ⌊np

⌋/q rectangles R(p, pq), and no more, can fit in

R(m, n).


Proof. By Theorem 172 exactly k =⌊

mp

⌋ ⌊np

⌋squares R(p, p) can fit in

R(m, n), this means at most kq of R(p, pq) can fit in R(m, n).

All we need to show is a tiling that uses kq tiles.

Consider the rectangle R(p bm/pc , p bn/pc) that fits in R(m, n).For this rectangle, a tiling exists by Theorem 156. This tiling uses k

q

tiles. Thus we can fit kq of R(p, pq) in R(m, n).

Theorem 175. Let {R(p, pq), R(pq, p)}. Then

GT (R(m, n)) = mn−⌊

mp

⌋ ⌊np

⌋p2.

[Not referenced]

rectangles 165

Proof. From Theorem 174 we can fit⌊

mp

⌋ ⌊np

⌋/q rectangles, and since

each has an area of p2q, the total covered area is at most⌊

np

⌋p2 and

therefor the gap is at least GT (R(m, n)) = mn−⌊

mp

⌋ ⌊np

⌋p2.

6.1.3 Tiling by Rectangles of other Regions

In this section, we extend some of the results of the previous sectionto tiling of arbitrary simple figures by rectangles.

Theorem 176 (Csizmadia et al. (2004), Corollary 2.4). If a simply-connected region is tileable by bars of length k, the region has at least oneside divisible by k.


Theorem 177 (Csizmadia et al. (2004), Corollary 2.3). If a simply-connected region is tileable by m × n and n × m rectangles, then it musthave at least one edge whose length is divisible by m, and at least one edgewhose length is divisible by n.

[Not referenced]

Proof. Note that the region is tileable by both bars of length m, andby bars of length n. Therefor, by Theorem 176 one side is divisible bym, and one side is divisible by n.

For example, a region is not tileable by dominoes if all the edgeshave odd length.

These theorems are generalizations of Theorem 156 and Theorem159.

Problem 50. Is the following true or false? If a simply-connected figureis tileable by a set of rectangles T that can be partitioned into two sets T1

and T2 such that all the rectangles in T1 has widths with factor r and allrectangles in T2 has heights with factor s, then the figure has either an edgewith factor r, or an edge with factor s.

6.2 Order

Figure 152: Polyomino with order 10.

A polyomino that can tile a rectangle is called rectifiable.The rectangular order of a polyomino is the smallest number k

such that k copies of the polyomino tiles a rectangle. The odd rectan-gular order of a polyomino is the smallest odd number k such that kcopies of the polyomino can tile the rectangle, if such a tiling exists.Polyominoes that has an odd order are called odd, otherwise theyare called even (Klarner, 1965).4 The square order is the minimum 4 This notion of even is different from

the one in Section 4.3.4.

166 polyominoes

number of tiles required to tile a square.

6.2.1 Order Theorems

Theorem 178. If a polyomino P has order n, then a scaled copy of thepolyomino P′ order has order n.

[Not referenced]

Proof. Suppose P′ has order n′ < n, and that the associated rectangleis R′. If we divide each tile into S = k × k squares, we have a tilingof R′ by S. But this tiling is unique (Theorem 3), and is the trivialtiling with all squares lying in a grid. But if this is the case, we canscale the tiling by k, and so find a valid polyomino tiling of a rect-angle using n′ tiles, which means the order of P is smaller than n, acontradiction.

It follows that the order of P′ must be larger than or equal to n.But by scaling a tiling of R by P, we can find a tiling of R′ by P′.Therefor, the order of P is n.

Theorem 179 (Klarner (1969), Theorem 2). An unbalanced polyominowith 2n cells is even.


Proof. Since the polyomino has 2n cells, the rectangle must have aneven area (Theorem 1), and is therefor balanced (see Theorem 52).

But an odd number of unbalanced polyominoes with an evennumber cannot have an equal number of white and black cells, so aregion tiled by these cannot be balanced. Therefor, the only possibil-ity for a balanced region be tileable by these polyominoes must betileable by an even number of them.

Theorem 180 (Klarner (1969), Theorem 4). Suppose the columns of apolyominoes are alternately colored black and white. Suppose the polyominohas k black cells or k white cells regardless of its position and orientation.Then the polyomino is even.

[Not referenced]

Proof. Suppose R is a rectangle tiled by the polyomino. Since the areaof the polyomino is even, so must the are of the rectangle (Theorem1). WLG assume then the orientation of the rectangle is such that therectangle has the same number of black and white cells; that is, thewidth is even.

Then form two sets, B with the copies of the polyomino coloredwith k black cells, and W with the copies of the rectangle with k

rectangles 167

white cells. Clearly, |B| = |W|, and therefore the total number ofpolyominoes must be even.

The hull of a polyomino is the smallest rectangle that can fit it.The hull of a rectangle is the rectangle itself.

Theorem 181. A rectifiable polyomino must cover at least one corner of itshull.


Proof. There is no way to place the polyomino to cover a corner cellof the rectangle.

This theorem is also a consequence Theorem 130, which states thatany reptile must cover at least one corner of its hull.

Theorem 182 (Klarner (1965), (i) p. 18). If a polyomino has symmetryindex 2 or less, and is rectifiable, it is a rectangle.

[Not referenced]

Proof. The polyomino must cover at least one corner of its hull (Theo-rem 181).

Suppose the polyomino is in class All, Rot2 or Axis2. Then, bysymmetry, must cover all corners of its hull. If we place the poly-omino in a corner of a rectangle, we cannot have gaps between therectangle edge and the polyomino edge, therefor, all the cells betweentwo adjacent corners must be part of the polyomino. This must betrue for all edges of the polyomino, so we have a rectangle, possiblywith some holes. But if there is a hole, it is too small to fit even onepolyomino, so we cannot have holes. Therefor, the polyomino is arectangle.

Suppose the polyomino is in class Diag2. If it covers two adjacentcorners of its hull, by symmetry it covers 4 corners of the hull and thesame argument as above applies.

Figure 153: A polyomino (in blue) inDiag2 that covers only 2 corners of itshull.

Figure 154: A polyomino (in blue) inDiag2 that covers only 2 corners of itshull.

If it does not cover two adjacent corners (Figure 153), by symmetryit must cover two opposite corners. If this polyomino tiles a rectan-gle, the tiles in the bottom corners must have different orientations(there is only one way to place a tile in each corner). If we considerthe bottom edge, then there must be at least a pair of adjacent tileswith opposite orientations also adjacent to the rectangle edge. Thehulls of these tiles cannot overlap, and so the cells corresponding tothe hull corners that are not covered cannot be covered by anothertile (See Figure 154). Therefor, this arrangement is impossible, andtherefor this case cannot occur.

168 polyominoes

In all cases, the only possibilities are for polyominoes that arerectangles.

Theorem 183 (Klarner (1965), (ii) p. 18). If a polyomino fits inside arectangle, and covers two diagonally opposite corners of its this rectangle,and it tiles this rectangle, it is a rectangle.

[Not referenced]

Proof. We have two cases: either the polyomino demarcates twodisconnected sections, or it also covers a third corner of the hull. Inboth cases, we cannot fit the polyomino in the remaining regions tobe tiled; so therefor, the only possibility is that the polyomino is arectangle.

Theorem 184 (Reid (2014), p. 117). If a polyomino tiles a (different) oddpolyomino with an odd number of tiles, it is odd.

[Not referenced]

Problem 51. Does the converse hold?

Theorem 185. An order 2 polyomino must cover at least 2 adjacent cornersof its hull.


Proof. Any dissection of a rectangle into two connected pieces musthave at least one tile with two adjacent corners of the rectangle. Be-cause if this is not the case, then pairs of opposite corners must be-long to the same tile (Figure 6.2.1). But there is no way that the twoblue corners can be connected and the two pink corners can be con-nected. Furthermore, all the corners of the rectangle of a piece mustlie on that piece’s hull.

Figure 155: No dissection of a rectangleinto two pieces can make pairs ofopposite corners belong to the samepieces.Figure 6.2.1 shows it is possible for a order-2 polyomino to only

cover 2 corners of its hull.

Figure 156: An example of a order-2polyomino that covers only two of itshull’s corners.

Theorem 186. A polyomino that covers exactly two opposite corners of itshull, with the bottom one on the left, can only tile a rectangle if it and a copyrotated 90° counter-clockwise can tile a straight edge.


rectangles 169

Proof. There are up to eight possible orientations for a polyomino.Let’s call a polyomino with a covered corner on the bottom left left,and right otherwise. Suppose the polyomino tiles a rectangle, andconsider the polyominoes making the bottom edge of the rectan-gles. Since the corners of this edge must be filled, we must have a leftpolyomino on the left and a right polyomino on the right. And so, onthis edge, we must have at least one pair of left and right polyomi-noes adjacent.

The right polyomino cannot be the left polyomino reflected aboutthe Y-axis. If this was the case, the top covered corners would beadjacent, and so we could not covered the uncovered corners, and sothere would be a gap on the edge.

By similar reasoning, the right polyomino cannot be the left poly-omino reflected about the Y-axis and then rotated 180°.

Next, suppose the right polyomino is the left polyomino rotated90° clockwise.

If they are to pack an edge, their convex hulls must overlap, be-cause if they don’t, there is no way to tile the uncovered hull corners.

The center of rotation must lie within this overlapping hull.Now consider the two corners of the overlapping rectangle along

the edge. The left one must belong to L, (otherwise it would coverR’s hull), and therefor the right cell must belong to R. If we rotate theleft cell by 90° counter clockwise about the center of rotation, it over-laps with the right cell, which cannot be by Theorem 20. Therefor,this configuration is impossible.

That leaves only one possibility, namely, the right polyomino isrotated 90° counter-clockwise. Figure 6.2.1 shows that in this config-uration it is possible for the two polyominoes to pack an edge, andindeed, the polyomino can tile a rectangle.

Figure 157: The only way in which apolyomino that covers only 2 oppositecorners of it’s hull can pack an edgewith a copy. If such a polyomino cantile an edge, it has order 4.

Theorem 187. If a n-omino and a 90° rotated copy pack an edge of lengthk ≤√

n, the polyomino has order at most 4. And the order is exactly fourwhen the polyomino covers exactly two opposite corners of the hull.

[Not referenced]

Proof. By Theorem 20 we can add two more copies, rotated 180°and 270°degrees around the same center of rotation, and they won’toverlap. In this compound figure, the packed edge must also existin all four rotations; and together these form the border of a tiledrectangle (possibly with holes). So the four copies pack a rectangle(possibly with holes). Since the edge of length k, the rectangle musthave area k2 ≤ n. So there cannot be any holes, and in fact k2 = n.

170 polyominoes

Therefor, four copies tile a rectangle (with no holes), and so the orderof the polyomino is at most 4.

If the polyomino covers exactly two corners of its hull, it cannotbe a rectangle, so its order cannot be 1. It can also not have order 2

(since a order 2 polyominoes must at least cover two adjacent cornersof its hull by Theorem 185).

A gun is a bar graph with height 2, which we will draw upsidedown so that it resembles a gun (Figure 158).5 A gun has a left and 5 The terminology is from Dahlke,

http://www.eklhad.net/polyomino/gun.html.He does not give this exact definition.

right end. If the corners are not missing from an end, it is called ablunt end. A gun with one blunt end is called a blunt gun (Figure159); a gun with two blunt ends is a degenerate gun (Figure 160).The longest non-blunt end is called the barrel. If the other end is alsonon-blunt, it is called the tab. The side with the contiguous row ofcells is called the back of the gun; the opposite side is called the face.A contiguous band of cells of the face is called a tooth, and the spacebetween two teeth is called a gap. The number of cells in the tooth atthe blunt end of a gun is the thickness of the blunt end. A gun thatis a rectangle with two corners removed is called a solid gun (Figure161).

Figure 158: A gun is a bar polyomimowith height 2.

Figure 159: A blunt gun.

Figure 160: A degenerate gun.

Figure 161: A solid gun.

Theorem 188 (Dahlke). If a polyomino with height 2 is rectifiable, it is agun.6

6 This is (1) from the Gun Theorem,http://www.eklhad.net/polyomino/gun.html.


Proof. A polyomino of width 2 must cover at least two corners of itshull.

If there are exactly two corners covered, and they are opposite,then there are only two ways to put the polyomino in a corner. Eitherway, we have gaps along the two rectangle edges that can only befilled in one of up to two ways; an all cases new gaps form. No mat-ter how we continue along the edge, there are gaps, and so we cannever fill an adjacent corner. Therefor, the polyominoes cannot tile arectangle, so this case is not possible.

http://www.eklhad.net/polyomino/gun.html

http://www.eklhad.net/polyomino/gun.html

rectangles 171

If these are the only two, or all four corners are covered, then thecells between them must be covered, as can be seen by placing the tilein the corner of the rectangle. We have a bar of height 2.

Figure 162:

Suppose there are three corners covered, with the bottom rightcorner missing, and some squares from the top row. The first tilemust cover the origin in exactly this orientation. The barrel points tothe right. If it has more than two cells, it forces a second copy, rotated180°, to fit in the space under the barrel. Since there are squaresmissing from the top, this isolates a hole that cannot be tiled (Figure162).

Figure 163:

Figure 164:

If the barrel has length 1, and the gun is missing the second cellfrom the top row, the second gun can fill the hole under the barrelwith its blunt end.

The gun at (0, 2) must be vertical, otherwise it creates holes withthe first tile. If the barrel is up, and the third cell in the top row of thegun is covered, this causes an untileable hole (Figure 163). If the thirdcell is not missing, we can only cover (1, 1) by a vertical gun reflectedand moved down and right. But both copies occupy (1, n)(Figure164). Whatever the case, the barrel must point down and fills the gapat (1, 1).

Figure 165:

Figure 166:

But then either (2, 1) or (2, 2) cannot be covered without isolatinguntilable holes (Figures 165 and 166).

Therefor, if three corners are covered, there cannot be any cellsmissing from the top row, and so we have a bar.

Theorem 189 (The Gun Theorem, Dahlke). The only rectifiable poly-ominoes with width 2 and order higher than 2 are the T-tetromino, theY-pentomino, the Y-hexomino, the D-hexomino, and the heptomino B(1 · 22 ·12), except perhaps B(1 · 23 · 13).


Proof. The proof is long. For full details, see the reference. Here is anoutline:

In this proof, with "gun" we mean a gun that does not have order2. Let n be the length of the gun.

(1) Theorem 188.

(2) If two guns are adjacent with their ends on the floor, they mustbe back-to-back.

(3) Three guns cannot be adjacent if they are all vertical. Therfore,we cannot fill 3 adjacent cells with guns with tabs. We can atmost fill 4 cells with blunt guns.

172 polyominoes

(4) Gaps with width n > k ≥ 3 and height 2 cannot be filled withguns with a tab, and gaps with width n > k ≥ 5 and height 2

cannot be filled with blunt guns.

(5) Degenerate guns with two thick ends are not rectifiable.

(6) No thick ends can cover the origin.

(7) Degenerate guns with one thick end are not rectifiable.

(8) Degenerate guns with no thick ends are not rectifiable.

(9) Two guns aimed at the origin must cover (1, 1).

(10) Guns with tabs of length larger than one are not rectifiable.

(11) In a tiling of a rectangle by blunt guns, the barrel cannot touchthe blunt end,

(12) the blunt ends cannot touch, and

(13) the blunt edge cannot touch the barrel.

(14) Blunt guns that have barrels with more than one cell are notrectifiable.

(15) Blunt guns are not rectifiable.

(16) Solid guns are not rectifiable, except for the T-tetromino andthe D-hexomino.

(17) A gun with a barrel of length 1 and teeth of lenth more thanone are not rectifiable.

(18) A gun with a barrel of length 1 is not rectifiable.

(19) Guns with gaps are not rectifiable.

(20) Guns with a face of one cell are not rectifiable, except for theT-tetromino, Y-pentomino and Y-hexomino.

(21) Guns with a barrel with two cells are not rectifiable, except forthe D-hexomino.

(22) Guns with a barrel with more than two cells are not rectifiable,except perhaps for B(1 · 23 · 13) shown in Figure 167.

Figure 167: B(1 · 23 · 13), the onlypolyomino with width 2 that we do notknow whether it tiles a rectangle.(23) This covers all guns, except for B(1 · 23 · 13). It is not known

whether this gun is rectifiable.

Problem 52 (Open Problem). Determine whether B(1 · 23 · 13) (Figure167) tiles a rectangle.

rectangles 173

6.2.2 L-shaped polyominoes

After Reid (2014), we make the following definitions: Let L(a, b, c)be the polyomino which is made from two rectangles, R(a, c) andR(b, c), aligned along an edge as shown in Figure 168.

Figure 168: The polyomino L(a, b, c) ismade from R(a, c) and R(b, c).

Two copies of these tiles the rectangle R(a + b, 2c), called the basicrectangle. We will assume that gcd(a, b, c) = 1. This does not lead toany loss of generality since a L(ad, bd, cd) is similar to L(a, b, c).

Theorem 190 (Klarner (1969)). L(a, 2a, b) is odd with odd order at most15.

[Not referenced]

Proof. Scale the tiling of R(5, 9) by the right tromino by (a, b) to finda new tiling of R(5a, 9b) or R(9a, 5b) by L(a, 2a, b).

This gives us an infinite family of polyominoes with odd rectan-gular order at most 15 (Golomb, 1966, p. 104). Figure 169 shows anexample where the tiling has been scaled by a factor of 3 horizontallyand 2 vertically. It is not necessarily true that these polyominoes haveodd order 15. For example, L(1, 2, 1) has odd order 11 (Figure 180).

Figure 169: A infinite family of poly-ominoes with odd order at most 15

can be found by stretching the righttromino.

Theorem 191 (Fletcher (1996) via Reid (2014)). L(2s, 3s, t) is odd if s isodd.

174 polyominoes

[Not referenced]

Theorem 192 (Fletcher (1996) via Reid (2014)). L(2s, 3s, 1) is odd if s iseven.

[Not referenced]

Theorem 193 (Fletcher (1996) via Reid (2014)). L(s, 4s, t) is odd if s isodd.

[Not referenced]

Theorem 194 (Fletcher (1996) via Reid (2014)). L(s, 4s, 1) is odd for alls.

[Not referenced]

Theorem 195 (Fletcher (1996) via Reid (2014)). L(s, 4s, 3) is odd for alls.

[Not referenced]

Theorem 196 (Jepsen et al. (2003) via Reid (2014)). For odd n, L(1, n−1, 1) tiles R(n + 2, 3n), and thus is odd with odd order at most 3(n + 2).

[Not referenced]

Theorem 197 (Jepsen et al. (2003) via Reid (2014)). For n ≡ 2(mod 4), L(1, n− 1, 1) is odd.

[Not referenced]

Polyomino OO Rect

L(1, 5, 1) 21ab 9× 14c

L(3, 4, 1) 33ab 11× 21c

L(4, 5, 1) 49ab 21× 21c

L(2, 3, 2) 33ab 15× 22c

L(1, 4, 2) 45ab 18× 25c

L(1, 5, 2) 35ab 20× 21a

L(2, 5, 1) 57a 9× 21c

L(2, 5, 2) 63a 18× 49a

L(2, 3, 3) 55a 11× 75 a

L(2, 7, 3) 133d 57× 63d

L(4, 5, 3) 119d 51× 63d

L(1, 8, 3) 125d 45× 45d

Table 29: Odd orders for various L-polyominoes. aMarshall (1997) bReid(1997) cReid dReid (2014)

Theorem 198 (Reid (2014)). Suppose a is odd, 2c divides b, and gcd(a, b, c) =1. Then L(a, b, c) tiles R((a+ b)(a+ 2b)+ b, 5c(a+ b)), and so is odd withorder at most 5((a + b)(a + 2b) + b).

[Not referenced]

Theorem 199 (Reid (2014)). If gcd(a + b, cd) = 1, then L(a, b, c) is odd.

[Not referenced]

Theorem 200 (Reid (2014)). If L(1, k− 1, 1) tiles R(m, n), then either mis even, or m ≥ k− 1, and the same for n.

[Not referenced]

rectangles 175

6.2.3 Known Orders

There are polyominoes of order 4k for any integer k. Table 30 summa-rizes results for other orders. There are no known polyominoes withrectangular order 6, but the tilings in Figures 170 and 209 suggestthat such polyominoes can exist.

Figure 170: A rectangular tiling usingsix 12-ominoes. The polyomino hasorder 2.

Order Tiling

1 All rectangles2 Figure 179

4 Figure 185

10a Figure 152

18a Figure 172(a)

24a Figure 172(b)

28a Figure 173

50a Figure 174

76a Figure 176

60b Figure 175

92a Figure 177

96a Figure 178

180c http://www.cflmath.com/Polyomino/8omino10_rect.html



Table 30: Examples of orders. FromaGolomb (1996, p. 97–100), bGrekov,http://polyominoes.org/rectifiable andcReid, http://www.cflmath.com/Polyomino/rectifiable_data.html.

Golomb (1996) gives a tiling of anoctomino with 312 copies, but thispolyomino was later found to haveorder 246 (Reid, http://www.cflmath.com/Polyomino/8omino11_rect.html).

http://www.cflmath.com/Polyomino/8omino10_rect.html



http://polyominoes.org/rectifiable

http://www.cflmath.com/Polyomino/rectifiable_data.html

http://www.cflmath.com/Polyomino/rectifiable_data.html



176 polyominoes

Figure 171: Marshall (1997).

(a) Polyomino of order 18. (b) Polyomino of order 24.Figure 172: Golomb (1996).


rectangles 177


178 polyominoes


rectangles 179


180 polyominoes


rectangles 181


182 polyominoes

Currently, we do not know which other orders greater than 4 arepossible, and in particular, we do not know if there are any polyomi-noes with an order that is odd other than 1 (Winslow, 2018, OpenProblem 5).

Only two known polyominoes has order equal to 6 modulo 8:(Reid, http://www.cflmath.com/Polyomino/11omino7_rect.html and http:

//www.cflmath.com/Polyomino/8omino11_rect.html).It is easy to come up with polyominoes of order 2: simply take a

rectangle with even area, and divide it into two with a lattice curvewith 180°-rotational symmetry (such a curve is called centrosymmet-ric).

Figure 179: An example of a polyominowith rectangular order 2.

Table 31 shows results for know odd orders. There are also somefamilies, for example odd orders of the form 3(p + 2) occur for allodd prime p (Reid, 1997).

Figure 180: Polyomino with odd order11.

Figure 181: A polyomino with oddorder 27.




rectangles 183



184 polyominoes

Odd Order Tiling

1 All rectangles11 Figure 180

15 R(5, 9) in Figure 203

21 R(7, 15) in Figure 209 and 213

27 Figure 181

33 Figure 182

35 Figure 183

45 http://www.cflmath.com/Polyomino/y5_rect.html


Table 31: Examples of odd orders. FromGolomb (1966), Grekov, Reid (2014).

We do know polyominoes cannot have order 3.

Theorem 201 (Stewart and Wormstein (1992)). If three congruent copiesof a connected polyomino P tile a rectangle, then P is itself a rectangle andthe tiling can only be one of the two shown in Figure 184.

Figure 184: The two ways in which arectangle can be disected into threecongruent shapes.


The proof of this theorem is long, but not very difficult. See the refer-ence for details.

Because rectangles have order 1, this theorem implies no poly-omino of order 3 exists.

There are 3 known ways in which polyominoes of order 4 can fittogether (Golomb, 1996, p. 89-99):

(1) In a construction with symmetry class Rot2.

(2) In a construction with symmetry class Axis2.

(3) In a construction with symmetry class Rot.

Figure 185: Examples of order 4 poly-ominoes.

Problem 53.

(1) Are all of the order 4 polyominoes of this type?

Most polyominoes that tile rectangles occur as families. Currently,there are families of order 2, 4, 8, 10, ....

Here are some known families:

(1) Families of order 4s (Golomb, 1996, pp. 102–204).

(2) Family of order 8 (Reid, 1998)7. The first three polyominoes is

7 The first polyomino in this family wasdiscovered by Marshall (1997), but Reid(1998) discovered the family it is part of.

shown in Figure 186.

(3) Families of order 2(a2 + b2) with gcd(a, b) = 1 (Marshall,1997). This family is constructed from a right triangle withsides a and b. Take b copies of a centrosymmetric curve , and

http://www.cflmath.com/Polyomino/y5_rect.html


rectangles 185

replace side of length a with b copies, and side of length bwith a copies. Figures 187-189 shows some examples.

(4) Families with odd order 3n + 6 for odd primes n (Marshall,1997). Examples are shown in... If n is prime, the R(n + 2, 3n)is minimal rectangle with a tiling. It is not know whetherthe rectangles are minimal for composite n, except for n =

9, 15, 21.

6.2.4 The orders of small polyominoes

The orders of small polyominoes is given in Table 32.

6.3 Prime Rectangles

A prime rectangle of a polyomino is a rectangle that can be tiledby that polyomino, and cannot be split into two smaller rectanglesthat can be tiled by that polyomino. A strong prime rectangle is arectangle that can be tiled by the polyomino, and cannot be tiled bysmaller rectangles that can be tiled by that polyomino (Reid, 2005, 3.1– 3.5).8 8 Some authors use the word "prime

rectangle" for what we refer to as a"strong prime rectangle" ((Klarner,1981)).

It should be clear that every strong prime rectangle is a primerectangle. Is the converse true? For sets of polyominoes, it is not.For example, I3 and I4 tile R(5, 5), but cannot do so with a fault.However, for single polyominoes we do not know.

Problem 54 (Reid (2005), Question 3.9). Is there a polyomino with aprime rectangle that is not a strong prime rectangle? This is an open prob-lem.

Theorem 202 (Reid (2005), Theorem 3.6). The set of prime rectangles ofa polyomino is finite.

[Not referenced]

See the reference for a proof. For two alternative proofs, see de Bruijnand Klarner (1975).

Theorem 203 (Reid (2005), Proposition 3.10 ). A rectangle has only oneprime rectangle: itself.9 9 Mentioned without proof in Klarner

(1981).

[Not referenced]

Proof. We need to show that any rectangle R(m, n) 6= R(p, q) that istileable by a rectangle P = R(p, q) can be split into two rectanglesthat can each be split into two rectangles, each of which is tileable byP.

186 polyominoes

Figure 186: A family of polyominoes oforder 8.

rectangles 187

Figure 187: Order 10


188 polyominoes

Polyomino O PR SO OO MR

Monomino 1 1 1 1 2

Domino 1 1 2 1 2

Bar 1 1 3 1 2

Right 2 2 12 15 2

Bar 1 1 4 1 2

Square 1 1 1 1 2

L 2 2 4 ∞ 2

T 4 1 4 ∞ 4

I 1 1 5 1 2

L 2 2 20 21 4

P 2 2 20 21 2

Y 10 40 20 45 9

1 1 6 1 2

1 1 6 1 2

2 ≥ 8 6 21 ?

2 ≥ 1 24 ? ?2 ≥ 1 24 ? ?2 ≥ 1 24 ? ?2 5 6 11 ?4 2 24 ? ?18 ≥ 26 96 ∞ ?92

a ≥ 45 ≤ 384 ? ?

1 1 7 1 2

2 2 28 27 6

2 ≥ 3 28 33 4

2 ≥ 4 28 ≤ 57 ?28 ≥ 37 28 45 ?76

b ≥ 26 ? ≤ 153 ?

1 1 8 1 2

1 1 2 1 2

2 ≥ 4 2 ? ?2 ≥ 1 2 ? ?2 ≥ 5 8 ? ?2 ≥ 5 8 ? ?

1 1 9 1 2

1 1 1 1 2

2 ≥ 2 36 ≤ 33 ?2 ≥ 4 4 15 ?4 ≥ 15 4 ≤ 87 ?

8 ≥ 2 12 ? ?

Table 32: Table giving order in-formation about polyominoes.Adapted from Grekov. See http://polyominoes.org/rectifiable.

O = OrderPR = Prime RectanglesSO = Square OrderOO = Odd OrderMR = Minimum ReptileaDahlke (1989b) bDahlke (1989a)



rectangles 189


From Theorem 160 we know that one of the following is true:

(1) p | m and q | n

(2) pq | m and n = px + qy for x, y > 0.

In the first case, we can divide the rectangle into either

• R(m− p, n) and R(p, n) if m > p

• R(m, n− q) and R(m, q) if n > q

In the second case, we can divide the rectangle in R(m, px) andR(m, qy), two rectangles of case (1).

So no other rectangle other than R(p, q) can be a prime of R(p, q).

6.3.1 Trominoes

Theorem 204 (Chu and Johnsonbaugh (1985)). The right tromino tiles arectangle iff 3|mn, and when one side is 3 the other is even.


Proof. (The proof follows the outline in Ash and Golomb (2004), p.48.)

Figure 190: The three ways a trominocan fit into the top left corner of arectangle.

(1) R(3m, 2n) divides into m columns and n rows of R(3, 2), whichis tileable by right trominoes.

(2) R(6k, 2n+ 3) can be split into one R(6k, 3) and several R(6k, 2n)rectangles. The first is tileable with 3k copies of R(2, 3), andthe second by (1).

190 polyominoes

(3) R(9 + 6k, 2n + 5) can be split into four rectangles: R(9, 5) (seeFig 203), R(6k, 2n + 5) (tileable by (2)), (9, 2n) (tileable by (1))and R(6k, n2) (tileable by (1)).

(4) R(3, 2n + 1) is not tileable. There are only three ways to placea tromino in the top left corner. Only two of them can work;for both these cases the only tiling option is to complete theR(3, 2) rectangle. This reduces the tiling to that of R(3, 2n− 1).We continue this, until a single row of three cells is left, whichis untileable.

Figure 191: Trying to tile R(3, 2n + 1)with right trominoes always result in acolumn that cannot be tiled.

6.3.2 Tetrominoes

Theorem 205 (Martin (1991), p. 50). The skew tetromino does not tile anyrectangle.

[Not referenced]

Proof. The proof follows directly from Thereom 186.Here is an alternative proof given in Martin (1991): The polyomino

can be placed in the corner in one of two ways, which are equivalent.The placement produces two notches which can fit the polyominoonly one way each, so that both neighbors are determined. Thisproduces new notches, which eventually result in a untileable cellin the adjacent corner of the rectangle.

Theorem 206 (Golomb and Klarner (1963)). The L-tetromino tilesR(m, n) iff 8 | mn and m, n > 1.

[Not referenced]

Proof. The L-tetromino satisfies the conditions of Theorem 179, andhence it is even. Therefor, the area must be divisable by 8. The proofthat all these rectangles have tilings uses the same ideas as used inTheorem 204. Figure 205 shows a tiling for R(8, 3).

We next want to prove that if a rectangle is tileable by T-tetrominoes,both sides are divisible by four. To do this, we will show that anytiling of the quadrant satisfy certain properties, and if we stack rect-angles together to form a quadrant tiling, these properties can onlybe satisfied if both sides are divisible by 4.

We introduce some terminology from (Walkup, 1965, Definition 1).An edge of the quadrant is called a cut if it coincides with an outeredge of some T-tetromino in every tiling of the quadrant. A vertexin the quadrant is called cornerless if it does not coincide with an

rectangles 191

outside vertex of a T-tetromino in any tiling of the quadrant. A vertexis a type-A vertex if it is congruent to (0, 0) or (2, 2) modulo 4. Avertex is a type-B vertex if it is congruent to (0, 2) or (2, 0) modulo 4.

Theorem 207 (Walkup (1965), Lemma 1). In a tiling of the quadrantby T-tetrominoes, every type-B vertex is cornerless, and every edge incidentwith a type-A vertex is a cut.


Proof. For integers k ≥ 0, let P(k) be the proposition that the theoremholds for all type-A and type-B vertices on or below the line x + y =

4k.P(0) is true, because the two edges incident on the origin are cuts.

We will prove that P(k) implies P(k + 1), and will use the Figure192 as reference, which shows the situation for k = 2. The cuts andcornerless points required for P(k) are shown by heavy lines anddots.

(1) We show that edges a and b, and their translates, are cuts.

If a tiling of the quadrant contains the tetromino 1− 2− 3−4, it will also contain 8 − 10 − 11 − 12, and so all upwardstranslates of 1− 2− 3− 4, because no other arrangement cancover cells 8 and 9. But this leads to two cells (13, 14) at the y-axis that cannot be covered. Therefor, the tiling cannot contain1− 2− 3− 4. By symmetry, it can also not contain 1− 5− 6− 7.

Figure 192:

There are four other ways to cover 1; all of them has a and bas outer edges, and therefor a and b are cuts, and so are theirtranslates. These are marked in Figure 2.

(2) We now show that point α is cornerless. Suppose, to the con-trary, that there is a tiling containing a T-tetromino having α asan outside corner.

In this tetromino the cell forming the outside corner at α can-not be 1, 2, or 3 because of the nearby cut segments. Conse-quently, either 1− 2− 3− 5 or 1− 2− 3− 6 is a tetromino ofthe supposed dissection.

But no tetromino could then contain square 6 or 5 respectively,as A is a cornerless point. This proves that α is cornerless. Bythe same or similar arguments all translates of α, either insidethe quadrant or on the axes, are cornerless.

(3) Finally, we show edges a and b, and their translates, are cuts.

192 polyominoes

By symmetry, it is enough to show a is a cut, and since c isa cut, it is enough to show that if an edge a is a cut, so is itstranslate d.

Suppose, a is a cut, and suppose, for a contradiction, thatcells 7 and 8 lie in the same tetromino. The tetromino cannotcontain 4, since α is cornerless. It must contain two of the threecells 9, 10, and 11. If it contained 9, there would be no way tocover 12 and 13. The tetromino must thus be 7− 8− 10− 11.The cells 14 and 15 must be covered by different tetrominoes,since γ is cornerless. Then the only way to cover 15 and 16 isfor 15− 17− 18− 19 to be part of the tiling. Then upwardstranslates of 15− 17− 18− 19 must be part of the tiling, whichmeans that there is no way to cover 20 and 21, a contradiction.Therefor, 7 and 8 must lie in different tetrominoes and so d is acut.

We have shown a and b and their translates are cuts.

Taken together, we have shown that P(k) implies P(k + 1), and thetheorem follows from induction.

Theorem 208 (Walkup (1965), Theorem 1). The T-tetromino tilesR(m, n) iff 4 | m and 4 | n.


Proof. If. The T-tetromino tiles R(4, 4); these can be stacked to formany rectangle R(m, n) = R(4m′, 4n′).

rectangles 193

Only if. We cannot have m congruent to 2 module 4, because thisputs a corner of R at a type-B vertex, which is impossible since type-B points are cornerless (Theorem 207). Moreover, m cannot be con-gruent to 1 or 3 module 4. If the edge of the rectangle is placed ateither P or Q, there is no way to cover cells 22 or 23. Thus, m and bysymmetry, must be congruent 0 modulo 4.

Problem 55 (Walkup (1965), Definition 2, Theorem 2). Let R =

R(m, n) be a rectangle with a tiling by T-tetrominoes. A block is a 2× 2square whose vertices have even coordinates. A chain of R is any minimalsubset of R that is both a union of blocks and a union of T-tetrominoes fromthe tiling.

Show that if we color every other block of R in the checkerboard fashion,then each tetromino has three cells in one block and one in an adjacentblock. Every chain consists out of an even number of blocks, which may becyclically ordered in such a way that the blocks are alternately colored blackand white, and the T-tetrominoes of the chain contain three cells of the oneblock and one cell of the succeeding block.

6.3.3 Pentominoes

We state a useful result for V-like n-ominoes with one leg length3 and the other leg length n − 2. We use the notation Vn for thesepolyominoes (Saxton Jr, 2015).

Theorem 209 (Saxton Jr (2015), Theorem 2.3, p. 10). For n ≤ 5, Vn doesnot tile a quadrant.


The proof is long and tedious with many cases. See the reference fordetails.

Theorem 210 (Cibulis and Liu (2001)). The F-, S-, T-, U-, V-, W-, X-,and Z-pentominoes do not tile a rectangle.

[Not referenced]

Proof. The X-pentomino does not cover any corner of its hull, so itcannot tile a rectangle 181.

Placing the F-, T- and Z-pentominoes in any position in a cornerleaves one cell that cannot be tiled.

The U-pentomino can cover the corner in only one way withoutcutting off any cells. But then the gap in the U can be covered onlyone way, which leaves one cell that cannot be covered.

The S- and W-pentominoes are not rectifiable by Theorem 186.

194 polyominoes

The V-pentomino cannot tile a quadrant (Theorem 209) and thusnot a rectangle (Theorem 133).

An alternative proof from Cibulis and Liu (2001): The V-pentominocan be placed in the corner in two ways. The first creates an inacces-sible pair. The second, forces a second to be placed as shown (other-wise it is not tileable for the same reason as before), and a third. Thisconfiguration leads to an inaccessible pair.

Theorem 211. The L-pentomino tiles:

(1) R(5m, 2n)

(2) R(10m, 2), R(10m, n + 4)

(3) R(15m, 7 + 2n)

[Not referenced]

Theorem 212. The P-pentomino tiles these and only these rectangles:

(1) R(5m, 2n)

(2) R(10m, 2), R(10m, n + 4)

(3) R(15m, 7 + 2n)

[Not referenced]

Proof. Figure 209 gives tilings of R(2, 5) and R(15, 7). From thiswe can construct R(5m, 2n) which proves (1). In particular, we canconstruct R(15, 2), so together with R(15, 7) we can construct allR(15, 7 + 2k), which proves (3). From (1) we can construct R(10, 5)and R(10, 2), putting these together proves (2).

Since 5|mn (Theorem 1), 5 divides either m or n. The only remain-ing rectangles we have not given tilings for are R(1, 5m), R(5, 2n + 1),and R(10, 3). We prove now these are impossible.

The pentomino cannot fit R(5, 1), so m, n > 1.Any rectangle that fits k polyominoes must also fit k R(2, 2)

squares.

R(5, 2n + 1) fits⌊

52

⌋ ⌊2n + 1

2

⌋= 2n squares (Theorem 172), but

2n + 1 are required; therefor no tiling of R(5, 2n + 1) is possible.Similarly, R(10, 3) fits only 5 squares, but 6 are required. So

R(10, 3) cannot be tiled.

Theorem 213 (Sillke (1992)10). The complete list of 40 prime-rectangles ofthe Y-pentomino is given in Table 34. 10 According to Reid (http://www.cflmath.

com/Polyomino/y5_rect.html) the firstcomplete list was published in Fogelet al. (2001), but I could not locatethe reference. The complete list wasidentified before then by Sillke (1992).Reid (2005, Example 5.2) gives anoverview of the publishing history ofthe prime rectangles.

[Not referenced]



rectangles 195

Proof. See Reid (2005, Example 5.2) for an outline.

(a) Checkerboard coloring (b) 2-coloring (c) 2-coloring

(d) 3-coloring (e) 2-coloring (f) 2-coloring

Figure 193: Various coloring used forproofs in this section.

6.3.4 Hexominoes

Hexomino Reason

A XC Thm. 189

D XE ×High F ×Low F ×G Thm. 188

H ×I X

Hexomino Reason

J XK ×L XM ×Long N Thm. 186

Short N Thm. 186

O XP XQ Thm. 186

Hexomino Reason

R XLong S Thm. 186

Long T ×Short T ××U XV Thm. 209

Wa ××Wb Thm. 186

Wc ×

Hexomino Reason

X Thm. 181

Italic X Thm. 181

High Y XLow Y Thm. 189

Short Z Thm. 186

Tall Z ×High 4 ×Low 4 ×

Table 33: Summary of proofs for pen-tominoes that do not tile rectangles.

Proving which hexominoes cannot tile rectangles is a bit tedious.In most cases we find an inaccessible cell within two steps, in othercases we can use one of the theorems we to show the polyomino isuntilable. This is summarized in Table 33.

11 http://www.cflmath.com/Polyomino/f6_rect.html).

Note that Reid calls this hexomino theF-hexomino.

Theorem 214 (Reid11). If the U-hexomino tiles R(m, n), then

http://www.cflmath.com/Polyomino/f6_rect.html

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196 polyominoes

(1) 12 | mn

(2) 4 | m or 4 | n

.

[Not referenced]

Proof. The U-hexomino is unbalanced, so the first part follows fromTheorem 179.

This implies that all rectangles are of the form R(4m, n) or R(2m +

2, 2n + 2). We prove that the R(2m + 2, 2n + 2) rectangle cannot betiled: Color the rectangle such that (x, y) is black if x and y are botheven. No matter how the polyomino is placed, it covers an even num-ber of black cells. However, R(2m + 2, 2n + 2) has an odd number ofblack cells. Therefor, it cannot be tiled.

12 http://www.cflmath.com/Polyomino/j6_rect.htmlTheorem 215 (Reid12). If the J-hexomino tiles R(m, n), then 4 | m or

4 | n.

[Not referenced]

Proof. The polyomino always tiles in pairs, as shown in Figure 194, sowe only need to show the tiling capability of these shapes. The areais therefor divisible by 12, and so it suffices to prove that we cannottile R(4m + 2, 4n + 2).

Figure 194: J-hexominoes always tile inpairs.

Apply this coloring: a cell is black if and only if x or y, but notboth, are divisible by 4 (Figure 193(e)). No matter how we place oneof the pairs, the pair always cover an odd number of black cells.R(4m + 2, 4n + 2) has an even number of black cells; however, itmust be tiled by an odd number of pairs, which will cover an oddnumber of black cells, which is a contradiction. Therefor, a tiling ofR(4m + 2, 4n + 2) is impossible, and so one side must be divisible by4.

13 http://www.cflmath.com/Polyomino/a6_rect.htmlTheorem 216 (Reid13). If the A-hexomino tiles R(m, n), then 4 | m or

4 | n.

[Not referenced]

Proof. The A-hexomino is unbalanced with an even number if cells,and so is even (Theorem 179). Thus the area is divisible by 12, and itsuffices to show that we cannot tile R(4m + 2, 4n + 2).

Apply this coloring: a cell is black if and only if x and y are botheven (Figure 193(b)). No matter how we place the polyomino, it al-ways cover an odd number of black cells. R(4m + 2, 4n + 2) has an

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rectangles 197

even number of black cells; however, it must be tiled by an odd num-ber of pairs, which will cover an odd number of black cells, which isa contradiction. Therefor, a tiling of R(4m + 2, 4n + 2) is impossible,and so one side must be divisible by 4.

14 http://www.cflmath.com/Polyomino/d6_rect.htmlTheorem 217 (Reid14). If D-hexomino tiles R(m, n), then 6 | m or 6 | n.

[Not referenced]

Proof. Apply this coloring: a cell is black if and only if x + b y3 c is

even (Figure 193(f)). No matter how we place the polyomino, it al-ways cover the same number of black and white squares. For rect-angles of the forms R(6m + 2, 6n + 3) and R(6m + 4, 6n + 3), wecan be colored so that the number of black and white cells are differ-ent; which means they cannot be tiled. The only other possibility isR(6m, n) or R(m, 6n), which means 6 divides on of the sides.

15 http://www.cflmath.com/Polyomino/d6_rect.htmlTheorem 218 (Reid15;Reid (2005), Theorem 5.12). If D-hexomino tiles

R(m, n), then 4 | m or 4 | n.

[Not referenced]

The proof is not very difficult, but requires some concepts that wehave not covered. See the second reference for details.

16 http://www.cflmath.com/Polyomino/g6_rect.htmlTheorem 219 (Reid16; Reid (2003), Theorem 5.4). If a G-hexomino tiles

R(m, n), then 4 | m or 4 | n.

[Not referenced]

As with the previous theorem the proof requires concepts we havenot covered. See the second reference for details.

17 http://www.cflmath.com/Polyomino/y6_rect.htmlTheorem 220 (Reid17). If the Y-hexomino tiles a rectangle with an odd

side, then the other side is divisible by 8.

[Not referenced]

Proof. R(2m + 1, 8n + k) is not tileable for odd k, since at least oneside must be even (Theorem 1). If R(2m + 1, 8n + 2) is tileable, thenso is R(2m + 1, 8n′ + 4); if R(2m + 1, 8n + 6) is tileable, then so isR(2m + 1, 8n′ + 2)., then so is R(2m + 1, 8n′′ + 4).

Therefor, it is enough to show R(2m + 1, 8n + 4) is not tileable forany integers m and n.

Consider the numbering

f (x, y) =

1 if x and y are both even

−1 if x and y are both odd

0 otherwise.

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198 polyominoes

No matter how the tile is placed, it covers ±2, or 2 (mod 4).R(2n + 1, 8n + 4) also covers a total that is 2 (mod 4). However, itwould require an even number of tiles, which would cover a totaldivisible by 4; a contradiction.

Therefor, R(2m + 1, 8n + 4) is not tileable by Y-hexominoes, and soneither is R(2m + 1, 8n + k) for any k not divisible by 8.

Therefor, if R(2m + 1, n) is tileable by Y-hexominoes, n is divisibleby 8.

6.3.5 Other polyominoes

Theorem 221 (Klarner (1969), Theorem 6). The P-octomino tiles R(m, n)iff m, n > 3 and 16 | mn.

[Not referenced]

Proof. If. If m, n > 3and 16 | mn, then R(m, n) can be cut into rectangles with dimensionsR(4, 4), R(5, 16), R(6, 8), and R(7, 16), which are all tileable as shownin Figure 195 (See Problem 56).

Figure 195: The prime rectangles of theP-octomino.Only if. If R(m, n) is tileable, it is obvious m, n > 3. To show that

mn is divisble by 16, we will show the P-octomino is even.Since 8 divides mn, 4 divides some side, say m. We apply the

coloring in shown in Figure 196 on the rectangle. The polyomino is oftype (a, b, c) if it covers a amber cells, b blue cells and c cherry cells.

Figure 196: F0,3, a 3-coloring.

Let k1···6 denote the number of tiles in the tiling with types (3, 4, 1),(1, 3, 4), (4, 2, 2), (2, 6, 0), (2, 2, 4), (0, 6, 2) respectively. It is easy tocheck no other types are possible.

rectangles 199

Since the total number of blue and cherry cells are equal, we have

3k1 + k2 + 4k3 + 2k4 + 2k5 = k1 + 3k2 + 2k3 + 4k5 + 2k6,

from which we get

k1 + k3 + k4 = k2 + k5 + k6.

The total number of tiles is k1 + k2 + · · · k6 = 2(k1 + k3 + k4), whichmeans the total is even. Hence the P-octomino is even, and hence anyrectangle R(m, n) that is tileable by this polyomino must have mndivisible by 16.

Problem 56. Show that any rectangle R(m, n) with mn divisible by 16 andm, n > 3, can be divided into the four rectangles R(4, 4), R(5, 16), R(6, 8),and R(7, 16).

6.3.6 Prime Rectangles of Small Polyominoes

200 polyominoes

Polyomino Prime Rectangles

3× 2 9× 5L 4× 2 8× 3T 4× 4P 5× 2 15× 7L 5× 2 15× 7Y 10× 5 20× 9 30× 9 45× 9 55× 9 14× 10 16× 10 23× 10

27× 10 20× 11 30× 11 35× 11 45× 11 50× 12 55× 12 60× 1265× 12 70× 12 75× 12 80× 12 85× 12 90× 12 95× 12 20× 1330× 13 35× 13 45× 13 15× 14 15× 15 16× 15 17× 15 19× 1521× 15 22× 15 23× 15 20× 17 25× 17 25× 18 35× 18 25× 22

23× 24 24× 29 24× 35 24× 41 24× 47 24× 53 24× 59 24× 6324× 65 24× 71 24× 77 24× 83 24× 89 24× 95 24× 101 24× 10224× 103 24× 107 24× 108 24× 113 24× 114 24× 119 24× 120 30× 6430× 68 30× 72 30× 80 30× 88 30× 92 30× 96 30× 100 30× 10430× 106 30× 108 30× 112 30× 116 30× 120 30× 124 · · · 32× 3632× 42 32× 48 32× 54 32× 60 32× 66 · · · 48× 48 · · ·· · ·3× 4

4× 6 5× 12

2× 7 11× 49 13× 35 19× 21

9× 12 9× 20 9× 28 12× 13 12× 14 12× 17 12× 19 12× 21

12× 24 12× 25 12× 29 15× 28 15× 32 15× 36 15× 40 15× 4415× 48 15× 52 16× 18 16× 27 16× 30 16× 33 16× 39 16× 4220× 21 20× 24

3× 4

2× 6 7× 12 8× 15 9× 14 9× 16 9× 34 10× 15 11× 18

3× 4

Table 34: Prime Rectangles of polyomi-noes

rectangles 201

6.4 Fault-free tilings

Recall that a fault in a tiling is a line on the grid (horizontal and verti-cal) that goes through the figure and is not crossed by any tile.

We are mostly interested in fault-free tilings because a fault-freetiling can not be broken down into two (or more) tilings by smallerrectangles.

Any prime rectangle is fault-free by definition. To find all possiblerectangles of a polyomino that has fault-free tilings, we make use offollowing

• The prime rectangles we looked at in the previous section.

• The theorems describing which rectangles have fault-freetilings by rectangles (described later in this section).

• Tiling extensions: systematic ways of building bigger rectan-gles from smaller ones. We will see several ways that fault-freerectangles can be extended.

• Finding more basic tilings of rectangles not covered by anyof the above or proving that they don’t exist. Once we havea basic rectangle, we also apply the three methods above toextend the set of rectangles with fault-free tilings.

Polyomino Fault-Free Rectangles

m× n, 2 | mn, m, n > 5, (m, n) 6= (6, 6)

m× n, 3 | mn, m, n > 3

I m× n, 4 | (p, q), p, q > 8

T 4m× 4nL 4k× 4k (2 + 4k)k (3 + 4k)× 8k (5 + 4k)× 8k

P m× n 5|mn if m = 2 then n = 5, if n = 2 then m = 5

L ?Y ?

Table 35: Fault-free Rectangles ofpolyominoes

6.4.1 Rectangles

Theorem 222. Tilings of a rectangle by a square cannot be fault-free.


202 polyominoes

Proof. Suppose the square has side k. Then the only rectanglestileable are R(mk, nk) for some integers m and n (Theorem 154). Atiling is given by putting m squares in each of n rows. By Theorem 3

this tiling is unique, and since it has a fault, no fault-free tilings of therectangle exists.

Theorem 223 (Robinson (1982), Theorem 4). Let 1 < m < n andgcd(m, n) = 1. The only rectangles with fault-free tilings of a m × npolyomino are these:

(1) (kmn+ pm+ qn)× lmn, with k ≥ 1, l ≥ 3, 0 < p < n, 0 < q < m(Figures 197 and 198)

(2) (kmn + pm)× (lmn + qn), with k, l ≥ 2, 0 < p < n, 0 < q < m(Figure 199)

(3) (kmn + pm)× (lmn), with k ≥ 2, l ≥ 3, 0 < p < n (Figures 199and 200)

(4) (kmn)× (lmn + qn), with k ≥ 3, l ≥ 2, 0 < q < m (Figure 198)

(5) (kmn)× (lmn), with k, l ≥ 3 (Figure 198)


To get some intuition of the this theorem, a useful exercise is toconstruct one of the minimal rectangles in Figures 197 –200 for agiven tile, and see if you can find the ways to extend it (withoutreferring to the figure). After this, you will be able to be able to seemore easily how the diagrams can be stretched for different tilings.

The following theorem is equivalent to the above, but often easierto work with.

Theorem 224 (Graham (1981), Theorem p.125). Let 1 ≤ m < n andgcd(m, n) = 1. Then a rectangle p, q has a fault free tiling if and only if:

(1) Each of m and n divides p or q,

(2) p and q can each be expressed as xm + yn in at least two ways, withx, y > 0, and

(3) for m = 1 and n = 2, p, q 6= 6.


We state a special case for bars:

Theorem 225. A rectangle R(p, q) has a fault-free tiling by bars R(1, m) ifand only if both the following hold

(1) m | p or m | q

rectangles 203

Figure 197: kmn × (mn + pm + qn),k ≥ 3, 0 < p < n, 0 < q < m

204 polyominoes

Figure 198: kmn × (mn + pm + qn),k, l ≥ 3, 0 ≤ p < n, 0 ≤ q < m. If q > 0,k may be 2.

Figure 199: (kmn + pm)× (lmn + qn),with k, l ≥ 2, 0 < p < n, 0 < q < m

rectangles 205

Figure 200: kmn × (2mn + pm), withk ≥ 3, 0 < p < n

(2) p, q ≥ 2m.


Proof. If. Figure 202 gives a tiling for rectangles that satisfy theconditions.

Only if. The first condition is necessary by Theorem 155.By Theorem 224, we must have p = x + ym in at least two ways.

This is possibly only if p > 2m, in which case we can write p =

x + m = x′ + 2m. The same holds for q. Therefore, we must havep, q > 2m.

Theorem 226 (Robinson (1982), Corollary 5 ). The minimal rectanglewith a fault-free tiling is a 3mn× (mn + m + n) rectangle18. 18 This was first proven for bars in

Scherer (1980)

[Not referenced]

6.4.2 Trominoes

The straight tromino is covered by Theorem 225, so we only dealwith the right tromino.

Theorem 227 (Aanjaneya and Pal (2006), Theorem 2). All p× q rectan-gles with p, q ≥ 4, 3 | pq, admit a fault-free tiling with trominoes.

206 polyominoes

Figure 201: A fault-free tiling of 13× 18rectangle by 2× 3 rectangles. The tilingwas obtained from Figure fig:ff-rect1with k = 3, p = 2, q = 1. Can you seehow we can reduce the width of thisrectangle and keep it fault-free? Whycan we not reduce the height?

Figure 202: A fault-free tiling of 3n×(2n + 1) rectangle by n× 1 rectangles,and an extension.

rectangles 207

[Not referenced]

See the reference for a proof.Since two trominoes can be put together to form a 2× 3-rectangle,

any rectangle that has a fault-free tiling with 2× 3 tiles will also havea fault-free tiling with right trominoes. Form Theorem 224 we knowthis is possible for rectangles with p, q either an odd number greaterthan 11 or an even number greater than 14 provided at least one isdivisible by three.

There are also rectangles with fault-free tilings not decomposableinto 2× 3 rectangles. The basic tromino fault-free tilings shown inFigure 203. These can be extended (similar to the extensions we havefor dominoes) to for all other rectangles that satisfy the conditions.

Figure 203: Basic fault free tilings usingtrominoes. Note that the gap is notreally a gap — the missing R(6, 9)is present in the top-right as R(9, 6).Image adapted from Aanjaneya and Pal(2006, Figure 4).

All the tetrominoes except the square admit fault-free tilings. ByTheorem 224 The I-tetromino has fault-free tilings for rectangles p, qwith p, q ≥ 9 and 4|(p or q). The smallest can be derived from Figure202.

6.4.3 Tetrominoes

The bar (Theorem 225) and square tetromino (Theorem 222) havealready been dealt with in the previous section. The skew tetrominoedoes not tile a rectangle, so here we will only consider the L and Ttetrominoes.

Theorem 228. There exist fault-free tilings by the T tetromino for anyrectangle that can be tiled by the T-tetromino (that is, for all 4m× 4n rect-angles).

[Not referenced]

208 polyominoes

Proof. Figure 204 shows a generic fault-free tiling. The blue sectionscan be repeated any number of times (including 0), leading to a ringthat can be filled with 4× 4 squares. This accounts for all 4m × 4nrectangles, which are the only ones that the T-tetromino can tile (The-orem 208).

Figure 204: A generic fault-free tilingwith T-tetrominoes.

Theorem 229. Fault-free tilings by L-tetrominoes exist for all rectanglesthat can be tiled by L-tetrominoes, except when m = 2 and n > 4 or n = 2and m > 4.

[Not referenced]

Proof. Fault-free tilings exist for the following rectangles (Figure 205):

(1) 4× 4

(2) 6× 4

(3) 3× 8

(4) 5× 8

Figure 205: The basic fault-free tilingsof rectangles by the L-tetromino.

These basic tilings can all be extended to give fault free tilings ofthe following:

rectangles 209

(1) 4k× 4k

(2) (2 + 4k)× 4k

(3) (3 + 4k)× 8k

(4) (5 + 4k)× 8k

Together, these classes account for all rectangles that can be tiled byL-tetrominoes except for rectangles with width or height 2. Of these,only the 2× 4 and 4× 2 are fault free.

The extensions of the first two classes can be done by insertingappropriate rectangles as shown in Figure 206.

Figure 206: The standard extensions ofthe 4m× 4n and 2 + 4m× 4n rectangles.

Vertical extensions of the other two classes is done in the sameway.

To extend the 3× 8k rectangle horizontally: take off one tile at thecorner, and insert the shape shown in figure, and add the tile in theopening (you will need to rotate it). Doing this twice is equivalent toinserting a cylinder; we therefor call this inserting a half cylinder.

Figure 207: To extend the 3× 8k rectan-gle, remove the pink corner tile. Addthe blue shape shown to fill the gap,and finally add the pink tile back tocomplete the rectangle.

To extend the 5 × 8k rectangle horizontally, add a basic 5 × 8rectangle, and flip the square shown in the figure.

6.4.4 Pentominoes

There are only 3 pentominoes to consider: the L-pentomino, the P-pentomino, and the Y-pentomino.

210 polyominoes

(a)

(b)

Figure 208: To extend the 5× 8k rect-angle, add another 5× 8 rectangle, andthen flip the orange square to removethe fault.

Theorem 230. The P pentomino have fault-free tilings of all rectanglesthat it can tile, except for rectangles with one dimension 2 when the other islarger than 5.

[Not referenced]

Proof. We know that one of the dimensions must be divisible by 5

(Theorem 1). It is not hard to show 5×m rectangles only have tilingsif m is even.

The P-pentomino can tile the following basic rectangles fault-free:

• 2× 5, 4× 5, 6× 5• 10× 7, 10× 9, 15× 7, 15× 9• 15× 11, 15× 11, 15× 13, 15× 15

See Figures 209 and 210.These can be extended to tile the following rectangles:

(1) 2m× 5n (if m = 1, then n = 1 too.)

(2) 2m + 5k× 10n

(3) 2m + 5× 5n + 5

Extension strategies are shown in Figure 211.This list make all rectangles with one side divisible by 5, except

5×m when m is odd, and rectangles of the form 2× 5n when n > 1,as is required.

Other strategies can also be used to extend basic rectangles. Figure212 shows an example where we put a 7 × 15 rectangle next to a2× 15 rectangle, and then eliminate the fault by retiling the orangetiles.

rectangles 211

Figure 209: Fault-free tilings of basicrectangles by the P pentomino.

212 polyominoes


rectangles 213



214 polyominoes

Theorem 231. The L-pentomino has these basic fault-free tilings (incom-plete):

• 10× 7• 15× 7• 15× 9


Figure 213: Basic fault-free rectangles.

Theorem 232. The L-pentomino has these fault-free tilings (incomplete):

(1) 10× (7 + 2k)

(2) (5k + 2p)× 5`


Proof.

(1) See Figure 214.

(2) The 2× 5 rectangle has a fault free-tiling, and so any fault-freetiling by R(2, 5) will also be fault-free if replace the rectangleswith the tilings by the pentomino. The fault-free tilings ofR(2, 5) is given by Theorem 223.

rectangles 215

Figure 214: Fault-free extensions.

Problem 57. Complete the lists of Theorems 231 and 232.

Finding all the rectangles with fault-free by Y-pentominoes is moretricky. Some extensions is shown in Figures: 215–222. Extension likethese are also given in Cibulis and Mizniks (1998).

Problem 58. Find all the fault-free tilings of rectangles by the Y-pentomino.

Figure 215: Example of an extension ofa fault-free rectangle. We can duplicatethe blue-purple region n times (includ-ing 0) to extend the rectangle vertically;as shown in the next three figures. Wecan also repeat the pink-purple regionto extend the rectangle horizontally.

216 polyominoes

Figure 216: Example of an extension ofa fault-free rectangle.



rectangles 217



218 polyominoes

Figure 221: Another way to extend the5× 10 rectangle. Place two copies nextto each other, and rotate the orange tilesas shown to eliminate the fault.


6.4.5 Extensions

We saw a few ways that rectangles can be extended:

• Through cylinder insertion.

• Through half-cylinder insertion.

• Through concatenation combined with fault-elimination (byretiling a subregion).

6.5 Simple Tilings

A tiling of a figure with more than two tiles is simple if no subsetof tiles (with more than one tile) form a rectangle strictly inside thefigure. For rectangles, simple tilings are also fault free (otherwise,the rectangle will split into two rectangles at the fault, at least one ofwhich must have more than one tile).

Theorem 233 (Martin (1991), p. 15). No rectangle has a simple tilingwith dominoes.

[Not referenced]

Proof. It is obvious for R(m, 1) or R(1, n).A rectangles R(m, n) with m, n > 1 have no peaks or holes, and

therefor any tiling of it has a flippable pair (which is a rectangle) assubtiling (Theorem 65), and is therefore not simple.

rectangles 219

Theorem 234 (Martin (1991), p. 15). Any simple tiling of the plane withdominoes is fault-free.

[Not referenced]

Proof. Suppose there is a fault, WLG let it be a horizontal fault. Thepattern on the fault must be alternating vertical and horizontal domi-noes, since any two adjacent dominoes of the same orientation willform a rectangle (and therefor the tiling would not be simple.) There-for, we can find an LR-configuration, which means there is a forcedflippable pair inside the induced pyramid (Theorem 96), whichmeans the tiling cannot be simple. Therefor, there can be no fault-line.

Theorem 235 (Martin (1991), p. 16). The plane has a unique (up toreflection) simple tiling with dominoes.

[Not referenced]

Proof. WLG, suppose the tiling contains a vertical domino, with 6

neighbors, labeled 1–6, starting from the bottom, going anticlockwise.

(1) In position 1, we cannot fit a vertical domino, so it must fit ahorizontal domino. Let’s say it lies to the right.

(2) Similarly, in position 4, we must fit a horizontal domino. Itcannot also lie to the right, since otherwise we will have an LRconfiguration, which will force a flippable pair (Theorem 96),therefor it must lie to the left.

(3) In position 2, we cannot fit a vertical domino (since it willeither overlap with the domino on 1, or form a rectangle),therefor it must be horizontal.

(4) A similar argument applies to position 5.

(5) In position 3, a horizontal domino will form a rectangle withthe domino at position 2, therefor it must be vertical.

(6) A similar argument applies to position 6.

We can now repeat the same argument for the new vertical domi-noes, except that because we have some neighbors placed, we do nothave a choice we had in step (1). And a symmetrical procedure canbe applied to horizontal dominoes; the entire tiling is forced.

If we made a different choice in step (1) above, we will constructthe same tiling, but reflected.

220 polyominoes

If we now consider general polyominoes, we can find simpletilings with 5 (Figure 231) and 7 (Figure 232) tiles, and as the fol-lowing theorem shows, for all numbers of tiles 7 or bigger.

Theorem 236 (Los et al. (1960)). There exists a simple tiling with n tilesfor all n ≥ 7.

[Not referenced]

Figure 223: A construction showinghow we can get a simple tiling for alln ≥ 5, n 6= 6. For example, to geta simple tiling with 7 tiles, place thefirst 7 tiles in the spiral as shown, andtruncate it at the dotted line.

Proof. Consider the construction shown in Figure 223. The rectanglesare placed in a spiral as shown. To get a simple tiling with n tiles,place the first n tiles in the spiral as shown, but truncate the last tileso that it does not extend beyond the rectangle. This works for alln ≥ 5, n 6= 6. (It does not work for n = 6, because truncating the sixthtile will make it the same size and aligned with the second tile.)

Proof. (Alternative proof (Chung et al., 1982, Theorem 1).) Using theblocks in Figure 224 we can construct simple tilings using any num-ber of tiles greater than 7 by choosing an appropriate left and rightblock, and any number (including 0) of center blocks. An exampleconstruction is shown in Figure 225.

Figure 224: Blocks that can be used toconstruct simple tilings using n piecesfor all n ≥ 7.

Theorem 237. There is no simple tiling with 6 rectangles.

rectangles 221

Figure 225: An example construction ofa simple tiling with 28 tiles.

[Not referenced]

Proof. (Adapted from Taxel (2016).)

(1) In a simple tiling, no rectangle can cover two corners of thecompleted figure. If there were such a piece, the remainingpieces would form a rectangle. Therefor, the four corners mustbe covered by different rectangles.

Figure 226:

(2) A corner piece must have at least 3 neighboring pieces. If acorner piece had two neighbors, they have to lie on oppositesides. If both neighbors were longer than the side, they mustoverlap. Therefor, at least one is the same size. But then itmakes a rectangle with the corner piece, which is not possiblesince the tiling is simple.

(3) Diagonally opposite corner pieces cannot touch. If they did,the remaining figure would be two rectangles. If either rectan-gle is a single piece, it forms a rectangle with one of the cornerpieces. Otherwise, it is tiled by smaller pieces, which violatesthe tiling being simple.

222 polyominoes

Now lets consider n = 6 specifically. Four of those six pieces mustbe in the four corners of the final figure (1).

Suppose the remaining two pieces are fully internal to the figure.Each of the outside pieces can only expose one side to the internalarea, so the internal area is rectangular. Filling it with the remainingtwo pieces creates a sub-rectangle with 2 pieces.

Figure 227:

Suppose on the other hand that all 6 pieces are on the boundary ofthe final figure, i.e. there are no internal pieces. So we have 4 cornerpieces and 2 edge pieces.

• Suppose a corner piece lies between two edge pieces. It musthave a third neighbor (2), but the only candidate is the diago-nally opposite corner, which violates (3).

Figure 228:

• Suppose the two edge pieces are adjacent. The four cornerpieces must be arranged so that the opening is in L-shape.There are two ways to fill the L-shape with rectangles; in eachcase one rectangle must be interior so both cannot be edgepieces and therefor this arrangement is impossible.19 19 The original proof omits this case.

• The only other arrangement for the edge pieces is on oppositesides of the final figure, say the left and right sides. The twotop corners are adjacent, cannot touch either of the bottomcorners, so the only way for them to have 3 neighbors is forboth corners to be adjacent to both edge pieces. This is notpossible.

Figure 229:

The last possibility is that we have 4 corner pieces, 1 edge piece,and 1 internal piece. The two corners next to the edge piece musthave the internal piece as their third neighbor. The edge piece hasthree internal sides and so must have at least three neighbors. Theonly possibility is that it is also adjacent to the internal piece. In asimilar argument to (2), the corners cannot be the same length asthe edge piece, and if both were longer then the edge piece and theinternal piece have matching lengths and form a rectangle.

Figure 230:

All possibilities lead to failure, so n = 6 is impossible.

rectangles 223

The average area of a tile in a tiling of a region R is given by|R|/n, where R is scaled so that the smallest tile has area 1, and nis the number of tiles20. 20 We always stretch the rectangle so

that the smallest tile is a square.

Figure 231: A simple tiling with 5 tiles.

Figure 232: A simple tiling with 7 tiles.

Figure 233: A tile that can be used toconstruct the tiling shown in Figurefig:rect-simple-plane

Theorem 238 (Chung et al. (1982), Theorem 2).

(1) Except for the simple tiling in Figure 231 that has average area 9/5,all other simple tilings of rectangles has average area stricter greaterthan 11/6.

(2) There is a unique simple tiling of the plane with average area of11/6. This tiling is shown in Figure 234.

[Not referenced]

Figure 234: A tiling of the plane withaverage area is 11/6.

6.6 Further Reading

Theorem 159 is given in a slightly more general form in the origi-nal paper (where it applies to n-dimensional bricks with arbitrarylengths, not necessarily integers). Wagon (1987) gives and discuss 14

proofs of this theorem, including some generalizations.Some problems relating to tiling rectangles by rectangles:

(1) Finding square tilings of squares. When each square has a dif-ferent area, the tiling is called perfect. Anderson (2013) gives achronology of squaring the square discoveries and is a goodintroduction to the problem.

(2) Finding whether a set of rectangles will pack a given rectangle. Asurvey of this is given by Lodi et al. (2002).

The question to show that there exists an integer C such thatR(m, n) is tileable by R(4, 6), R(6, 4), R(5, 7) and R(7, 5) for allm, n > C was asked in the The Fifty-Second William Lowell Putmanmathematical competition (Klosinski et al., 1992, Problem B-3). Narayanand Schwenk (2002) finds the minimum value C, and the thesis ofDietert (2010) considers which rectangles can be tiled by this set.

Barnes (1982a) and Barnes (1982b) considers tilings of bricks bybricks from a algebraic viewpoint. The n-dimensional version ofTheorem 171—the gap number bricks packed by rods —is provenin Barnes (1995). In Brualdi and Foregger (1974) the problem is dis-cussed for harmonic bricks, and in Barnes (1979) for sufficientlylarge" rectangles .

Frobenius numbers and related problems are covered in detail inRamírez-Alfonsín (2005).

Theorem 201 has been generalized by Maltby (1994) by showingthat trisecting a rectangle into 3 congruent pieces of any shape (not

224 polyominoes

necessarily polyominoes), implies the pieces are rectangles. Yuanet al. (2016) showed that if a square is dissected into 5 convex congru-ent pieces, there is a unique solution with rectangles only. (Unfortu-nately, this is not helpful for polyominoes, since the only polyomi-noes that are convex in the normal geometric sense are rectangles.)

Kenyon (1996) gives an algorithm for determining whether a rec-tilinear figure (not necessarily with all integer sides) is tileable withrectangles each of which has at least one integer side. One drawbackof this algorithm is that the tiling rectangles cannot be specified aspart of the input.

For more on tilings by bars, see Beauquier et al. (1995).Bloch (1979) contains tilings by rectangles for small rectangles.

Tables with the number of tilings for small polyominoes can be foundin Grekov (http://polyominoes.org/data).

For more on prime rectangles see de Bruijn and Klarner (1975),Klarner (1973), and Reid (2005). The latter also gives many examplesof finding prime rectangle for polyominoes and polyomino sets. Reid(2008) discusses tiling theorems for "sufficiently large" rectangles.

Korn and Pak (2004) consider T-tetromino chains and height func-tions of T-tetromino tilings. The number of tilings of rectangles byT-tetrominoes is considered in Merino (2008). Tilings of deficientrectangles by L-tetrominoes are discussed in Nitica (2004), and byT-tetrominoes in Zhan (2012). Gap numbers for the T-tetromino onrectangles is covered in Hochberg (2015).

http://polyominoes.org/data

7Plane Tilings

When and how polyominoes tile the plane are the central questionsof this section.

In many ways, plane tilings are more fundamental than tilings ofrectangles; for one thing — any tile set that tiles a rectangle can alsotile the plane.

Theorem 239 (Grünbaum and Shephard (1987), Chapter 3). In a tilingby any tile set T , there must be a tile with at most 6 neighbors.


A tiling is called monohedral if all the tiles in the tilesets are con-gruent1 (Grünbaum and Shephard, 1987, p. 20). 1 The terminology makes sense when

applied to tilings of other regions, butis generally only used to discuss planetilings.

Suppose there are two non-parallel non-zero vectors u and v suchthat U + u + v is the same tiling, then the tiling is periodic. If thereis only one such vector, the tiling is called half-periodic. If there is nosuch vector, the tiling is non-periodic.

Example 19. We can easily form half-periodic and non-periodic tilings ofthe plane from tiles using irrational numbers. Here are some examples usingdominoes:

• Form a half-strip using vertical and horizontal flippable pairs, usingalternate digits of π of each; that is, 3 vertical, 1 horizontal, 4 ver-tical, 1 horizontal, etc. Reflect the strip and form a complete strip;now stack these together to tile the plane. The resulting tiling ishalf-periodic.

• Form a strip using the same scheme as above; form another stripreversing the roles of vertical and horizontal flips. Now stack thesetogether using alternate digits of π for each.

Another method is to use a tiling with all dominoes vertical (and aligned).Flip a single flippable pair; this tiling is non-periodic. If we flip all the domi-noes in a row, we get a half-periodic tiling.

226 polyominoes

Some tile-sets can only tile the plane non-periodically. Such atile-set is called aperiodic. If the set is a single tile, we call the tileaperiodic. Currently, we don’t know if connected aperiodic tiles(polyominoes or non-polyominoes) exist.2 2 See Socolar and Taylor (2011) for an

example of a disconnected aperiodictile.

Informally, a fundamental region of a tiling is the smallest partof the tiling that we can translate to form the entire tiling. Moreformally, the fundamental region of a periodic tiling with period u, vis a set F of connected cells such that if x ∈ F, then x + mu + nv 6= Ffor all m 6= 0, n 6= 0. 3 3 The fundamental region is also called

a fundamental domain. Note, this isdifferent from the definition given infor example Grünbaum and Shephard(1987, p. 55), and (Kaplan, 2009, Defini-tion 3.1, p. 17) that is defined in termsof symmetry groups.

7.1 Tilings by Translation

A polyomino that can tile the plane by translation alone is calledexact (Beauquier and Nivat, 1990, Section 1). The smallest tiles thatare not exact are the pentominoes: F, U and T.

Theorem 240 (Translation Criterion, (Rhoads, 2005), Theorem 2). Atile admits a tiling of the plane if its border can be divided into six segments,A, B, C, D, E and F such that the pairs A − D, B − E and C − F aretranslates of each other. (Both edges in one of these pairs may be empty).

[Not referenced]

If word that can be written in the form ABCABC (possibly withone pair of factors empty), we call that form the BN-factorization ofthe word. A word can have more than one BN factorization.

Theorem 241 (Beauquier-Nivat Criterion, Winslow (2015)). A poly-omino is exact if its border has a BN factorization . .


7.2 Tilings by Translation and Rotation

Problem 59. Show that no The only Young diagrams that are exact arerectangles and L-shaped polyominoes.

Problem 60.

(1) Find the other factorizations of X5.

(2) Suppose P is a polyomino with border word x±1y±1x±1y±1 ·x±1y±1. Characterize the ones with a BN factorization..

Theorem 242. The families of polyominoes listed in table 36 are all exact.

[Not referenced]

plane tilings 227

Polyomino BN Factorization

X5 (xyx)(yx−1y) · (x−1y−1x−1)(y−1xy−1)

Z5 (x)(y)(x−1y2x−1) · (x−1)(y−1)(xy−2x)N5 (xy−1x2)(x)(y) · (x−2yx−1)(x−1)(y−1)

R(m, n) Rectangle (xm)(yn) · (x−m)(y−n)

A(n) Aztec diamond (xy)n(yx−1)n · (x−1y−1)n(y−1x)n

B(am · bn) L-shaped polyomino (yn−mxb)xayn · (x−bym−n)x−ay−n

W2k Even-area W-polyomino x[(xy)k−1x

]y · x−1

[(x−1y−1)k−1x−1

]y−1

W2k+1 Odd-area W-polyomino x(xy)ky · x−1(y−1x−1)ky−1

Yn General Y-polyomino x(yx3)(yx−1) · x−1(x−3y−1)(xy−1)

B(1k · 2m · 1n) xm(yxn)(yx−k) · x−m(x−ny−1)(xky−1)

Ck(a1, a2, · · · , ·am) Cylinder xk A · x−k A, where A = xkya1 xkya2 · · · xkyam

Table 36: BN factorizations for variouspolyominoes and families of polyomi-noes.Proof. The BN factorization is given in the table for each family,

which makes it exact (Theorem 241).

Note that this proves exactness for all n-ominoes with n ≤ 5,except for F,

Theorem 243 (Conway’s Criterion). A polygon admits a periodic tilingof the plane using only translations and 180° rotations if it has 6 points (atleast three of which are distinct), that satisfy these conditions:

(1) The boundary between A and B is congruent by translation to theboundary between E and D.

(2) The boundaries BC, CD, EF and FA are all centrosymmetric.

[Not referenced]

Problem 61. Show that the golygon with 8 sides (Figure 14) satisfies theConway criterion.

7.3 Isohedral and k-Isohedral Tilings

For two congruent tiles in a tiling there must be at least one rigidmotion of the plane that maps the one tile to the other. If this motionis also a symmetry of the plane, then this motion maps the wholetiling to itself, and the two tiles are transitively equivalent.

Transitive equivalence is a equivalence relation, that divides thetiles in the tiling into transitivity classes. If there are k classes, wecall the tiling k-isohedral4; if k = 1, we simply call the tiling isohe- 4 Also tile-k-transitive (Huson, 1993,

Section1, p.272).dral5.5 Also tile-transitive (Huson, 1993,Section1, p.272).

228 polyominoes

There are 93 kinds of isohedral tilings of the plane with markedtiles, distinguished by how a tile relates to its neighbors. 81 of thesecan be realized with tiles without markings. (Grünbaum and Shep-hard (1977), Grünbaum and Shephard (1987, 6.2.1)) 67 of the 93

tilings can be realized with polyominoes; 60 can be realized withpolyominoes without markings. The 81 kinds of isohedral tilings canall be described by 9 rules given in Table 37.6 6 According to Langerman and Winslow

(2015, Section 3) these are first shownin Heesch and Kienzle (2013, Table 10),and reproduced in Schattschneider andEscher (1990, p. 326).

Theorem 244. Each tile in an isohedral tiling has at most 6 neighbors.

[Not referenced]

Proof. By Theorem 239 there must be at least one tile with at most 6

neighbors. But since transformations that map any other tile to thisone maps the tiling to itself, it means all tiles must have at most 6

neighbors.

In general, a single tile can belong to different transitivity classesin a single tiling. Some examples are shown. Therefor, being k-isohedral is a property of the tiling and not the tile set. However, atile set that admits k-isohedral tilings, but no m-isohedral tilings form < k, is called k-anisohedral.

To summarize: tilings can be k-isohedral; tile sets can be k-anisohedral.There is k-anisohedral polyominoes for all k between 1 and 6

(shown in Figures 235-240). We do not know whether polyominoesexist for other k (Winslow, 2018, Problem 6).7 7 We do know that other tiles exist for

k = 8, 9, 10. See for example Myers.

Figure 235: An isohedral tile that doesnot satisfy Conway’s Criterion. (Myers)

plane tilings 229

Figure 236: An 2-anisohedral tile.Transitivity classes are shown in thesame color. (Myers)


230 polyominoes


plane tilings 231

Figure 239: An 5-anisohedral tile.Transitivity classes are shown in thesame color. This polyomino has thelargest known fundamental region(among polyominoes) that consists outof 20 tiles. (Myers)


232 polyominoes

Problem 62. Find the number of transitivity classes for the tiling in Fig241.

Figure 241: How may transitivityclasses does this tiling have?

In all the examples found, there is one tile per transitivity classin the fundamental region. Whether this is always the case is notknown. Only one example is known for a non-polyomino (shown inFigure 242).

There are essentially 9 ways a arbitrary tile can tile the plane, buttwo of these — Criteria 7 and 8 — cannot be realized with polyomi-noes, as the following theorem shows.

Theorem 245. An equilateral triangle cannot have all it’s vertices on latticepoints.

[Not referenced]

Proof. The area A of an equilateral triangle is given by A =√

34 s,

where s is the length of the side and given by s =√(a− a′)2 + (b− b′)2.

Therefor, the area is given by

A =

√3√(a− a′)2 + (b− b′)2

4.

This value is irrational if a, a′, b and b′ are all integers (see Problem63).

But by Pick’s Theorem (Theorem 14), the area of any polygonwith it’s vertices on the lattice is rational. This is a contradiction, andtherefor the equilateral triangle cannot have all its vertices on latticepoints.

Problem 63. Prove that if the sum of two squares is divisible by 3, it isdivisible by 9. (It follows that

√3 ·√

x2 + y2 cannot be ration is x and y arerational.)

plane tilings 233

Criterion Rules ConceptualDiagram

Polyomino Example

Criterion 1:Conway’sCriterion

(1) Sides a and d will be translates of each other.(2) Sides b, c, e, f will be centrosymmetric.

Criterion 2

(1) Sides a and d will be translates of each other.(2) Sides b and c will be centrosymmetric.(3) Sides e and f will glide reflect to each other.

Criterion 3

(1) Sides a and d will be translates of each other.(2) Sides b and c will glide reflect to each other.(3) Sides e and f will glide reflect to each other.

Criterion 4:TranslationCriterion

(1) Sides a and d will be translates of each other.(2) Sides b and e will be translates of each other.(3) Sides c and f will be translates of each other.

Criterion 5

(1) Sides a and d will be translates of each other.(2) Sides b and f will glide reflect to each other.(3) Sides c and e will glide reflect to each other.(4) The glide reflections above will be parallel.

Criterion 6

(1) Sides a and d will glide reflect to each other.(2) Sides b and f will glide reflect to each other.(3) Sides c and e will be centrosymmetric.(4) The glide reflections above will be perpendicular.

Criterion 7

(1) Side c will rotate 120° around D to side d.(2) Side e will rotate 120° around F to side f .(3) Side a will rotate 120° around B to side b.

Criterion 8

(1) Side c will rotate 120° around D to side d.(2) Side e will be centrosymmetric.(3) Side a will rotate 60° around B to side b.

Criterion 9

(1) Side b will rotate 90° around C to side c.(2) Side a will be centrosymmetric.(3) Side d will rotate 90° around E to side e.

Table 37: The 9 ways a tile can tile theplane.

234 polyominoes

n n-ominoes Holes Translation 180° Isohedral Anisohedral Non-tilers

1 1 0 1 0 0 0 0

2 1 0 1 0 0 0 0

3 2 0 2 0 0 0 0

4 5 0 5 0 0 0 0

5 12 0 9 3 0 0 0

6 35 0 24 11 0 0 0

7 108 1 41 60 3 0 3

8 369 6 121 199 22 1 20

9 1285 37 213 748 80 9 198

10 4655 195 522 2181 323 44 1390

11 17073 979 783 5391 338 108 9474

12 63600 4663 2712 17193 3322 222 35488

13 238591 21474 3179 31881 3178 431 178448

14 901971 96496 8672 85942 13590 900 696371

15 3426576 425449 16621 218760 43045 1157 2721544

16 13079255 1849252 37415 430339 76881 2258 10683110

17 50107909 7946380 48558 728315 48781 1381 41334494

18 192622052 33840946 154660 2344106 551137 7429 155723774

19 742624232 143060339 185007 3096983 93592 5542 596182769

20 2870671950 601165888 573296 9344528 2190553 18306 2257379379

21 11123060678 2513617990 876633 17859116 3163376 22067 8587521496

22 43191857688 10466220315 1759730 31658109 3542450 47849 32688629235

23 168047007728 43425174374 2606543 49644736 1065943 10542 124568505590

24 654999700403 179630865835 8768743 172596719 39341178 202169 475147925759

25 2557227044764 741123699012 10774339 228795554 31694933 28977 1815832051949

Table 38: Table showing counts for thenumber of various polyominoes that tilea certain way. From Myers.

plane tilings 235

Figure 242: An example of a 2-anisotropic tile where the transitiv-ity classes are not equally big. Thetiling was discovered by Myers, andBerglund.

kn 2 3 4 5 6

8 1 0 0 0 0

9 8 0 1 0 0

10 41 3 0 0 0

11 89 18 1 0 0

12 214 6 2 0 0

13 406 24 0 1 0

14 874 24 1 0 1

15 1107 49 1 0 0

16 2210 46 1 0 1

17 1316 60 2 0 3

18 7380 42 7 0 0

19 5450 85 2 0 5

20 18211 86 5 0 4

21 21866 199 2 0 0

22 47702 135 9 1 2

23 10390 149 3 0 0

24 201834 324 11 0 0

25 28784 182 8 1 2

Table 39: Table showing number ofpolyominoes that are k-anisohedralMyers.

236 polyominoes

n 180° as well Translation only

1 1 0

2 1 0

3 2 0

4 5 0

5 9 0

6 24 0

7 41 0

8 121 0

9 212 1

10 520 2

11 773 10

12 2577 135

13 3037 142

14 8081 591

15 13954 2667

16 32124 5291

17 41695 6863

18 118784 35876

19 150188 34819

20 411484 161812

21 604304 272329

22 1305265 454465

23 1954823 651720

24 5326890 3441853

25 7331606 3442733

Table 40: Polyominoes that tile theplane by translation Myers

plane tilings 237

7.4 m-morphic and p-poic Polyominoes

We may also ask in how many ways can a tile tile the plane. If weallow all transformations of a tile, a tile that is m-morphic tiles theplane in m distinct ways. If we only allow translations and rotations,we tile is p-poic if it tiles the plane in p distinct ways. If m = ∞, thenwe call the tile polymorphic; if p = ∞, we call the tile polypoic. Weknow tiles for all values of m ≤ 10 (Martin, 1991, p. 96).

Table 41 gives m and p values for the various polyominoes that tilethe plane.

(a) (b) (c)

(d) (e) (f) (g)

(h) (i) (j)

Figure 243: Diagrams of the polyomi-noes of Table 41.

Theorem 246. A polyomino is hypermorphic when it tiles two prime stripsof different width, or one strip in two different ways, or the period of thestrip is larger than 1.

[Not referenced]

This means interesting cases involve polyominoes that belongcharacteristically to PP (or AP if such polyominoes exist).

238 polyominoes

Tile m t p h

B(1 · 2 · 12 · 2) 1 0 0 12

Figure 243 (a) 1 0 0 14

B(1 · 2 · 12 · 2 · 1) 1 1 1 16

Figure 243 (b) 1 1 1 18

Figure 243 (c) 2 1 1 ∞B(1 · 2 · 13 · 2) 2 1 1 10

Figure 243 (d) ∞ 1 1 12

Figure 243 (e) 1 2 1 13

B(12 · 4 · 1) 2 2 2 14

Figure 243 (f) 3 2 2 ∞B(2 · 12 · 3) 3 2 2 6

Figure 243 (g) 4 2 2 12

Figure 243 (h) ∞ 2 2 13

Z(2, 1, 1, 2) 2 3 2 16

B(1 · 2 · 1 · 3) 3 3 3 19

Figure 243 (i) 4 3 3 ∞Z(2, 3, 1, 1) ∞ 3 3 14

B(1 · 5 · 1) 2 4 2 18

Figure 243 (j) 4 4 4 ∞B(3 · 2 · 10) 5 4 4 8

Tile m t p h

B(2 · 3 · 4 · 55) 6 4 4 12

B(4 · 5 · 66) 7 4 4 14

B(4 · 5 · 67) 8 4 4 16

B(43 · 23 · 83) 9 4 4 18

B(1 · 2 · 4 · 52) ∞ 4 4 ∞B(1 · 85 · 45) 5 5 5 10

B(43 · 14 · 94) 6 5 5 12

B(4 · 21 · 6) 7 5 5 13

B(22 · 12 · 72) 7 5 5 14

B(2 · 12 · 42) ∞ 5 5 ∞B(3 · 2 · 3) 3 6 3 6

B(44 · 284 · 214 · 144 · 74) 6 6 6 12

B(44 · 154 · 104 · 54) 7 6 6 13

B(22 · 102 · 52) 8 6 6 16

B(33 · 83 · 43) 10 6 6 19

Z(22, 1, 1) ∞ 6 6 ∞B(12 · 2 · 3 · 4) 7 7 7 14

B(72 · 142 · 82) 9 7 7 18

B(43 · 14 · 64) ∞ 7 7 ∞B(4 · 5 · 4) 4 8 4 8

Table 41: Examples of m-morphic andp-poic polyominoes. From Martin(1986).

(a) An 8-morphic polyomino. (b) A 9-morphic polyomino. (c) A 10-morphic polyomino.

(d) A 10-morphic poly-omino.

(e) A 11-morphic polyomino.Figure 244: Examples of polymorphicpolyominoes. The first three is fromWinslow (2018, Figure 8), the last twofrom Myers.

plane tilings 239

Figure 245: The 6 tilings of the Z-pentomino if reflection is not allowed.

240 polyominoes

mn 1 2 3 4 5 6 7 8 9 10 11 ∞ Skipped

1 1 0 0 0 0 0 0 0 0 0 0 0 0

2 0 0 0 0 0 0 0 0 0 0 0 1 0

3 0 0 0 0 0 0 0 0 0 0 0 2 0

4 0 0 0 0 0 0 0 0 0 0 0 5 0

5 1 0 0 0 0 0 0 0 0 0 0 11 0

6 0 0 0 0 0 0 0 0 0 0 0 35 0

7 6 6 3 0 0 0 0 0 0 0 0 89 0

8 20 18 4 2 0 0 1 0 0 0 0 298 0

9 193 84 14 6 1 0 0 0 0 0 0 752 0

10 749 257 41 2 2 1 0 0 0 0 0 2018 0

11 3222 809 148 31 12 3 2 0 0 1 0 2392 0

12 9026 1440 153 22 4 0 0 0 0 0 0 12803 1

13 25090 3645 435 94 25 4 3 3 0 0 0 9368 2

14 63746 5681 416 56 17 2 1 0 0 0 0 39176 9

15 180669 11842 665 85 13 1 0 1 0 0 0 86306 1

16 366557 16758 1128 142 37 6 2 0 0 0 0 162257 6

17 683157 30733 1473 264 63 9 2 1 0 0 0 111321 12

18 2192816 65557 2226 238 31 1 1 1 0 0 0 796444 17

19 2936540 72811 2130 360 96 7 2 1 1 1 1 369160 14

20 9444080 126363 3550 322 68 42 1 1 0 0 0 2552238 18

21 18299457 221446 4767 458 90 14 0 0 2 0 0 3394939 19

Table 42: The number of polyominoesthat are m-morphic Myers. (Meyersoriginally classified the monomino asmonomorphic—presumably because heallows non-discrete tilings.)

plane tilings 241

7.5 Non Tilers

We call a polyomino that does not tile the plane a non-tiler. Thesmallest non-tilers have 7 cells; there are only 3 without holes, shownin Fig. 246. The 20 simply-connected octominoes that don’t tile theplane is shown in Fig. 247. Numbers for non-tiles of polyomninoeswith 25 cells or less is shown in Table 38.

Figure 246: Heptominoes without holesthat don’t tile the plane.

Figure 247: Octominoes without holesthat don’t tile the plane.

When a tile does not tile the plane, we may wonder: how close canit come to tiling it?

One way to make this question precise is to ask “How many ringsof a tile can be place around another tile so that there are no holes?”

242 polyominoes

The answer to this question is called a tile’s Heesch number. Cur-rently, we don’t know of any polyomino that does not tile the planewith a Heesch number greater than three; in fact, we do not knowof any tile with Heesch number bigger than 5 (see Mann (2004) forexamples).

Figure 249 shows all the known polyominoes with Heesch number2, with some example tilings in Figure 248. The polyomino in Figure250 has Heesch number 3; is derived from the 4-polypillar in Mann(2004) and Mann and Thomas (2016).

Some results for polyominoes are shown in Table 43.

Heesch numbern Hole-free non-tilers 0 1 2 Unclassified

7 3 1 2

8 20 6 14

9 198 77 120 1

10 1390 770 620

11 9474 6029 3409 26 10

12 35488 29114 6369 5

13 178448 152339 26025 40 44

14 696371 642259 53537 26 549

Table 43: Heesch Numbers for Polyomi-noes Kaplan (2017)

A U-frame polyomino is a polyomino of the form x−hygx− f y−exdycx−by−z.

Theorem 247 (Fontaine (1991)). A U-frame polyomino has Heesch num-ber 2 if:

(1) a = g + h− e

(2) b = g + h

(3) c = h

(4) d = 2g + 3h

(5) f = g + h,

where 0 < g < e < h < 2e,

[Not referenced]

7.6 Aperiodic Tilings

While it is easy to find tilings that are non-periodic, it is not so easyto find tile aperiodic tile sets. Two examples of aperiodic sets areshown in Figures 251 and 253. The tiling of the first set is shown inFigure 252.

plane tilings 243

(a) Kaplan (2017)

(b) Rhoads (2003)

(c)

Figure 248: Some almost-tilings ofpolyominoes with Heesch number 2

244 polyominoes

Figure 249: Polyominoes with Heeschnumber 2. List obtained from Kaplan(2017).A Wang tile8 is a square with colored edges; in a legal tiling of

8 Golomb (1970) calls this a MacMahontile, after Major P.A. MacMahon whostudied their properties in MacMahon(1921). The question whether aperiodicsets of Wang tiles exist was first askedin Wang (1961). The first aperiodictile set is a set of 20,426 Wang tiles,described in Berger (1966), where theterm Wang tile is also used for the firsttime.

Wang tiles colors must match. Despite their simplicity, there are setsof Wang tiles that tile the plane aperiodically. Such a set must use atleast four colors Chen et al. (2014), and have at least 11 tiles Jeandeland Rao (2015). An example is shown in Figure 258.

There is a way to convert between polyomino problems and Wangtile problems (Golomb, 1970). To convert from Wang tiles to poly-ominoes, the basic idea is to use a base tile that is monomorphic, andencode the colors in binary by perforating the edges (by adding andremoving cells). For this to work, the base tile must be big enough.Figure 256 shows the sequence of Gi of base tiles introduced. To en-code 4 colors, we need the perforatable edge to be 2 cells wide, sowe can use G2. The set of 11 Wang tiles realized as polyominoes isshown in Figure 259.

Another encoding scheme (from Yang (2014)) is shown in Figure257.

To convert from polyominoes to Wang tiles, we will have a uniqueWang tile for each cell in the set. We choose one color for outeredges, and one color for each internal edge. This way, the only tilingswill match a polyomino tiling exactly.

Because of their simple structure, Wang tiles are much easier towork with. Because of the conversion, some general theorems thatcan be proven for Wang tiles also apply to Polyominoes.

Here are some examples:

plane tilings 245

Figure 250: A 99-omino with Heeschnumber 3.

Figure 251: An aperiodic tile set discov-ered by Roger Penrose derived from aset by Robert Amman(Penrose, 1994,Fig. 1.3, p. 32).

246 polyominoes

Figure 252: A patch of a plane tiling bythe Penrose set.

Figure 253: An aperiodic set discoveredby Matthew Cook (Wolfram, 2002, p.943).

plane tilings 247

Figure 254: An aperiodic set given in(Winslow, 2015, Fig. 5), modified from asimilar set in Ammann et al. (1992, Fig.1).

Figure 255: A tiling by Wilson’s tiles.

Figure 256: Golomb’s Encoding ofWang tiles.

248 polyominoes

Figure 257: Yang’s encoding of Wangtiles.

Figure 258: The smallest aperiodic set ofWang tiles.

plane tilings 249

Figure 259: Encoded as polyominoes.

(1) Whether a set of Wang tiles (polyominoes) can tile the plane isundecidable.

(2) Whether a set of Wang tiles (polyominoes) can tile an infinitestrip is decidable.

(3) There are no aperiodic tilings of a infinite strip.

7.7 Further Reading

Plane tilings for more general tiles than polyominoes is widely cov-ered. (Toth et al., 2017, Chapter 3) is a handy reference for terminol-ogy, main results and open problems. I already mentioned Grün-baum and Shephard (1987), which is the most comprehensive sourceon the topic. Kaplan (2009) is a lighter read, and especially suitable ifyou want to use the computer to experiment. Conway et al. (2016) isa very interesting book, focusing on symmetry, algebra, and topology,not only on the plane, but also on the sphere and hyperbolic plane.Other resources include Horne (2000), and a short survey Goodman-Strauss (2016). Open problems is listed and briefly discussed in Brasset al. (2005).

250 polyominoes

The 1270 types of 2-isohedral tilings of the plane is given in Del-gado et al. (1992). Huson (1993) contains information about classifi-cation of isohedral, 2-isohedraL and 3-isohedral tilings of the plane,sphere, and hyperbolic plane.

8Various Topics II

The purpose of this section is to introduce various interesting topicsand provide some references. I may expand some of these topicslater.

8.1 Compatibility

Two polyominoes P and Q are compatible if there is a finite1 re- 1 There are more interesting problemsif we remove this restriction: whatpolyominoes are infinitely compatiblebut not finitely? Can the infinite regionsthat serve as common multiples becharacterized?

gion R that they can both tile Cibulis et al. (2002). Of course, if onepolyomino tiles another, they are compatible. Any two rectanglesare compatible, since R(m, n) and R(m′, n′) both tile R(mm′, nn′) byTheorem 154. If P tiles P′ and Q tiles Q′, and P′ is compatible withQ′, then P is compatible with Q. It follows then that any two rectifi-able polyominoes are compatible. The monomino is compatible withany polyomino, a property not shared even with dominoes, whichis incompatible with . Minimal incompatible figures can be found inSicherman. There are many unproven conjectures for this problem.The compatibility of pentominoes is shown in Tables 44 and 45.

F I L N P T U V W X Y Z

F 1 10 2 2 2 2 4 4 2 2 2 2

I 10 1 2 2 2 4 12 4 10 × 2 20

L 2 2 1 4 2 2 2 2 2 44 2 2

N 2 2 4 1 2 2 2 2 2 16 2 2

P 2 2 2 2 1 2 2 2 2 4 2 2

T 2 4 2 2 2 1 4 2 14 4 2 2

U 4 12 2 2 2 4 1 2 2 × 2 4

V 4 4 2 2 2 2 2 1 6 × 2 4

W 2 10 2 2 2 14 2 6 1 ? 2 4

X 2 × 44 16 4 4 × × ? 1 2 ?Y 2 2 2 2 2 2 2 2 2 2 1 2

Z 2 20 2 2 2 2 4 4 4 ? 2 1

Table 44: Table showing the leastnumber the number of tiles necessary totile the LCM (as far as is known). FromSicherman.

SelfOverall compatibility unknownOverall incompatible

252 polyominoes

F I L N P T U V W X Y Z

F 1 10 2 2 2 2 4 6 2 2 2 2

I 10 1 2 2 2 32 ? 10 10 × 2 ?L 2 2 1 4 2 2 2 2 2 × 2 2

N 2 2 4 1 2 2 2 2 2 16 2 2

P 2 2 2 2 1 2 2 2 2 4 2 2

T 2 32 2 2 2 1 ? 2 16 4 2 30

U 4 ? 2 2 2 ? 1 ? 2 × 2 ?V 6 10 2 2 2 2 ? 1 6 × 2 4

W 2 10 2 2 2 16 2 6 1 ? 2 10

X 2 × × 16 4 4 × × ? 1 2 ?Y 2 2 2 2 2 2 2 2 2 2 1 2

Z 2 ? 2 2 2 30 ? 4 10 ? 2 1

Table 45: Table showing the leastnumber the number of tiles necessary totile the LCM (as far as is known). FromSicherman.

SelfOverall compatibility unknownHoleless compatibility unknownOverall incompatibleHoleless incompatible.Overall best solution has holes.

Also of interest is the set of regions that can be tiled by each of aset of polyominoes, called a common multiple; a smallest region thatis a common multiple of some polyominoes is called a smallest com-mon multiple (Cibulis et al., 2002). The smallest common multiple ofthe tetrominoes is shown in Figure 260.

Figure 260: The smallest (known)common multiple of the tetrominoes.

The basic idea is in a slightly more general setting in Golomb(1981). For more, see Barbans et al. (2003), Cipra et al. (2004), Pegget al. (2009), and Liu (2018b). Large lists of compatibility tables can befound on Resta and Sicherman.

various topics ii 253

8.2 Exclusion

What is the least number of monominoes we can place inside R(m, n)so we cannot place a polyomino P anywhere? What fraction of cellsof the plane must be covered by monominoes so that a polyomino Pcannot be placed anywhere? This problem is introduced in Golomb(1996, Chapter 3).

Theorem 248. If P is an efficient polyomino, it can be excluded from theplane with 1/|P| monominoes.

[Not referenced]

Proof. Suppose K is an optimal coloring for P. Then put a monominowherever K(x, y) = 0. Since however we place a polyomino it mustcover each color once, if there is no cells with color 0 available, it isimpossible to place P anywhere.

Among others, this gives us colorings for various rectangles (allthose whose one side divides the other, including squares and bars)by Theorem ??.

Problem 64. Is the converse true; that is: if we can exclude a polyomino Pfrom the plane with 1/|P| monominoes, is P efficient?

The following allow us to get a lower bound on the exclusionfraction. Suppose a certain arrangement excludes P, and P ⊂ Q.Then Q is also excluded by the same arrangement.

Problem 65. Find exclusion patterns for the hexominoes.

nm 1 2 3 4 5 6 7 8 9

1 1 2 3 4 5 6 7 8 9

2 2 4 5 8 8 12 12 16 16

3 3 5 9 10 13 18 18 18 27

4 4 8 10 16 17 17 25 32 32

5 5 8 13 17 25 26 26 26 41

6 6 12 18 17 26 36 37 37 37

7 7 12 18 25 26 37 49 50 50

8 8 16 18 32 26 37 50 64 65

9 9 16 27 32 41 37 50 65 91

Table 46: Inverse frequency for rectan-gles. (Adapted from Klamkin and Liu(1980))

This exclusion problem for pentominoes on rectangles has beendiscussed in Gravier and Payan (2001) and Gravier et al. (2007), andextended to other types of graphs. Klamkin and Liu (1980), Barnesand Shearer (1982) treats the exclusion problem on the plane.

254 polyominoes

(a) Excludes X, I (F5,2) (b) Excludes L, T, U, Y (F4)

(c) Excludes F, N, P, W (S2) (d) Excludes V (F13,8)

(e) Excludes Z(F3,0) Figure 261: Exclusion patterns forpentominoes. The pattern can beobtained from the coloring in bracketsby putting a monomino at each cellwith color 0. From Golomb (1996, Fig.67), Klamkin and Liu (1980, Fig. 3).

various topics ii 255

8.3 Fountain sets

The monomino tiles any polyomino. Are there other sets of polyomi-noes that tile “almost all” polyominoes?

The set in Figure 262 tiles all polyominoes, except for the monomino.

Figure 262: A fountain set. Thesepolyominoes can tile all polyominoes,except the monomino.

In general, a fountain set is a set of polyominoes that can tile allpolyominoes formed from appending a monomino to a member ofthe setKenyon and Tassy (2015).

Fountain sets are often infinite. In fact, any fountain set that con-tains no bars is infinite.

We can extend this idea in several ways:

• What smallest sets of polyominoes can tile all but a finite setof polyominoes? polyominoes that have area divisible by n?polyominoes that are balanced for some coloring K?

• What sets can tile all elements of the set appended with adomino? some other polyomino? some other set of polyomi-noes?

256 polyominoes

Bibliography

Aanjaneya, M. and S. P. Pal2006. Faultfree Tromino Tilings ofRectangles. arXiv preprint math/0610925.(Cited on pages 205 and 207.)

Alhazov, A., K. Morita, and C. Iwamoto2010. A Note on Tatami Tilings (Math-ematical Foundation of Algorithmsand Computer Science). (Cited onpage 123.)

Ammann, R., B. Grünbaum, and G. C.Shephard1992. Aperiodic tiles. Discrete & Com-putational Geometry, 8(1):1–25. (Cited onpage 247.)

Anderson, S. E.2013. Compound Perfect SquaredSquares of the Order Twenties. arXivpreprint arXiv:1303.0599. (Cited onpage 223.)

Ardila, F. and R. P. Stanley2010. Tilings. The Mathematical Intelli-gencer, 32(4):32–43. (Cited on pages 24

and 39.)Ash, J. M. and S. W. Golomb

2004. Tiling deficient rectangles withtrominoes. Mathematics Magazine,77(1):46–55. (Cited on page 189.)

Bar-Yehuda, R. and E. Ben-Hanoch1996. A linear-time algorithm forcovering simple polygons with sim-ilar rectangles. International Journalof Computational Geometry & Applica-tions, 6(01):79–102. (Cited on pages 17

and 18.)Barbans, U., A. Cibulis, G. Lee, A. Liu,

and B. Wainwright2003. Polyomino number theory (II). InMathematical Properties of Sequences andother Combinatorial Structures, Pp. 93–100. Springer. (Cited on page 252.)

Barnes, F. W.1979. Packing the maximum numberof m × n tiles in a large p × q rectan-gle. Discrete Mathematics, 26(2):93–100.(Cited on pages 163 and 223.)

Barnes, F. W.1982a. Algebraic theory of brick pack-ing I. Discrete Mathematics, 42(1):7–26.(Cited on page 223.)

Barnes, F. W.1982b. Algebraic theory of brickpacking II. Discrete Mathematics, 42(2-3):129–144. (Cited on page 223.)

Barnes, F. W.1995. Best packing of rods into boxes.Discrete Mathematics, 142(1-3):271–275.(Cited on page 223.)

Barnes, F. W. and J. B. Shearer1982. Barring rectangles from the plane.Journal of Combinatorial Theory, Series A,33(1):9–29. (Cited on page 253.)

Bassino, F., F. Becker, N. Bédaride, O. Bo-dini, C. Boutillier, B. D. Tilliére,T. Fernique, P. Guillon, A. Hashemi,É. Rémila, and G. Theyssier2015. Tilings and Tessellations.https://lipn.univ-paris13.fr/~fernique/isfahan/lectures_notes.pdf. (Cited on page 24.)

Beauquier, D. and J.-C. Fournier2002. Groups and tilings. Theoreticalcomputer science, 281(1-2):81–97. (Citedon page 122.)

Beauquier, D. and M. Nivat1990. Tiling the Plane with One Tile. InSymposium on Computational Geometry,Pp. 128–138. (Cited on page 226.)

Beauquier, D., M. Nivat, E. Rémila, andM. Robson1995. Tiling figures of the plane withtwo bars. Computational Geometry,5(1):1–25. (Cited on pages 18, 48, 49, 53,54, 56, 59, 69, and 224.)

Berger, R.1966. The undecidability of the dominoproblem, number 66. American Mathe-matical Soc. (Cited on page 244.)

Berglund, J.. A tile that is unbalanced - withdifferent sized transitivity classes.http://www.angelfire.com/mn3/anisohedral/unbalanced.html (Accessed 11 April2019). (Cited on page 235.)

Berlov, S. and K. Kokhas2004. Domino Tilings. 16-th summerconference International mathemati-cal Tournament of towns. (Cited onpage 122.)

Black, M.1947. Critical thinking. An introductionto logic and scientific method. (Cited onpage 27.)

Bloch, C. J.1979. Catalogue of Small RectangularPlans. Environment and Planning B: Plan-ning and Design, 6(2):155–190. (Cited on

page 224.)Bodini, O. and T. Fernique

2006. Planar dimer tilings. Com-puter Science–Theory and Applications,Pp. 104–113. (Cited on page 122.)

Bodini, O. and M. Latapy2003. Generalized tilings with heightfunctions. Morfismos, 7(1):47–68. (Citedon page 122.)

Bodini, O. and J. Lumbroso2009. Optimal partial tiling of Man-hattan polyominoes. In Discrete Geom-etry for Computer Imagery, Pp. 79–91.Springer. (Cited on pages 45, 69, 106,and 107.)

Bondy, J. A. and U. S. R. Murty1976. Graph theory with applications,volume 290. Macmillan London. (Citedon page 68.)

Bougé, L. and M. Cosnard1992. Recouvrement d’une pièce trapé-zoïdale par des dominos. Comptesrendus de l’Académie des sciences. Série 1,Mathématique, 315(2):221–226. (Cited onpage 49.)

Bousquet-Mélou, M., A. Guttmann, W. Or-rick, and A. Rechnitzer1999. Inversion relations, reciprocityand polyominoes. Annals of Combi-natorics, 3(2-4):223–249. (Cited onpage 45.)

Bower, R. J. and T. Michael2004. When Can You Tile a Box WithTranslates of Two Given RectangularBricks? the electronic journal of combi-natorics, 11(1):N7. (Cited on pages 155

and 157.)Brass, P., W. O. Moser, and J. Pach

2005. Tiling Problems. Research Prob-lems in Discrete Geometry, Pp. 161–182.(Cited on page 249.)

Brualdi, R. A.2009. Introductory combinatorics, 5

edition. Pearson. (Cited on page 123.)Brualdi, R. A. and T. H. Foregger

1974. Packing boxes with harmonicbricks. Journal of Combinatorial The-ory, Series B, 17(2):81–114. (Cited onpage 223.)

Bruckheimer, M. and A. Arcavi1995. Farey series and PickâAZs areatheorem. The Mathematical Intelligencer,17(4):64–67. (Cited on page 20.)

Butler, S., P. Horn, and E. Tressler2010. Intersecting domino tilings. TheFibonacci Quarterly, 48:114–120. (Citedon pages 97 and 98.)

Caspard, N., M. Morvan, E. Remila, andE. Thierry2003. Lattices of Tilings and Stability.

https://lipn.univ-paris13.fr/~fernique/isfahan/lectures_notes.pdf

https://lipn.univ-paris13.fr/~fernique/isfahan/lectures_notes.pdf

http://www.angelfire.com/mn3/anisohedral/unbalanced.html

http://www.angelfire.com/mn3/anisohedral/unbalanced.html

258 polyominoes

Technical Report 2003-25, Laboratoirede l’Informatique du Parallélisme,École Normale Supérieure de Lyon.(Cited on page 122.)

ccorn2017. Counting pairs of bits.Mathematics Stack Exchange.https://math.stackexchange.com/q/570546(version: 2013-11-17). (Cited onpage 18.)

Chavanon, F. and E. Remila2006. Rhombus tilings: decompositionand space structure. Discrete & Compu-tational Geometry, 35(2):329–358. (Citedon page 42.)

Chavanon, F., E. Rémila, et al.2003. Contractions of octagonaltilings with rhombic tiles. Techni-cal Report 2003-44, Laboratoire del’Informatique du Parallélisme, ÉcoleNormale Supérieure de Lyon. (Cited onpage 42.)

Chen, H.-H., W.-G. Hu, D.-J. Lai, and S.-S.Lin2014. Nonemptiness problems ofWang tiles with three colors. TheoreticalComputer Science, 547:34–45. (Cited onpage 244.)

Chu, I. and R. Johnsonbaugh1985. Tiling boards with trominoes.J. Rec. Math, 18:188–193. (Cited onpage 189.)

Chung, F., E. Gilbert, R. Graham,J. Shearer, and J. van Lint1982. Tiling rectangles with rectangles.Mathematics Magazine, 55(5):286–291.(Cited on pages 220 and 223.)

Cibulis, A.2001?a. Pentamino. I dal,a.http://nms.lu.lv/wp-content/uploads/2015/11/Pentamino1.pdf. (Cited on page 24.)

Cibulis, A.2001?b. Pentamino. II dal,a.http://nms.lu.lv/wp-content/uploads/2015/11/Pentamino2.pdf. (Cited on page 24.)

Cibulis, A. and A. Liu2001. Packing Rectangles with theL and P Pentominoes. Math Hori-zons, 9(2):30–31. (Cited on pages 193

and 194.)Cibulis, A., A. Liu, and B. Wainwright

2002. Polyomino number theory (I).Crux Mathematicorum, 28(3):147–150.(Cited on pages 251 and 252.)

Cibulis, A. and I. Mizniks1998. Tiling rectangles with pentomi-noes. Latv. Univ. Zinat. Raksti, 612:57–61.(Cited on page 215.)

Cipra, B., E. D. Demaine, M. L. Demaine,

and T. Rodgers2004. Tribute to a Mathemagician. CRCPress. (Cited on page 252.)

Ciucu, M.1997. Enumeration of perfect matchingsin graphs with reflective symmetry.Journal of Combinatorial Theory, Series A,77(1):67–97. (Cited on page 123.)

Ciucu, M.2003. Perfect matchings and perfectpowers. Journal of Algebraic Com-binatorics, 17(3):335–375. (Cited onpage 123.)

Cohn, H., R. Kenyon, and J. Propp2001. A variational principle fordomino tilings. Journal of the AmericanMathematical Society, 14(2):297–346.(Cited on page 68.)

Colomo, F., A. G. Pronko, andA. Sportiello2018. Arctic curve of the free-fermionsix-vertex model in an L-shaped do-main. arXiv preprint arXiv:1807.07549.(Cited on page 123.)

Conway, J. H., H. Burgiel, andC. Goodman-Strauss2016. The symmetries of things. AKPeters/CRC Press. (Cited on page 249.)

Conway, J. H. and J. C. Lagarias1990. Tiling with polyominoes andcombinatorial group theory. Journal ofcombinatorial theory, Series A, 53(2):183–208. (Cited on page 69.)

Csizmadia, G., J. Czyzowicz, L. Gasieniec,E. Kranakis, E. Rivera-Campo, andJ. Urrutia2004. On Tilable Orthogonal Poly-gons. INTERNATIONAL JOURNAL OFPURE AND APPLIED MATHEMAT-ICS., 13:443–460. (Cited on pages 16

and 165.)Csizmadia, G., J. Czyzowicz, L. Gasieniec,

E. Kranakis, and J. Urrutia1999. Domino tilings of orthogonalpolygons. (Cited on pages 36 and 37.)

Dahlke, K.. Tiling Rectangles With Polyominoes.http://www.eklhad.net/polyomino/index.html.(Cited on pages 24, 170, and 171.)

Dahlke, K. A.1989a. A heptomino of order 76. J.Comb. Theory, Ser. A, 51(1):127–128.(Cited on page 188.)

Dahlke, K. A.1989b. The Y-hexomino has order 92.Journal of Combinatorial Theory Series A,51(1):125–126. (Cited on page 188.)

Davey, B. A. and H. A. Priestley2002. Introduction to lattices and order.Cambridge university press. (Cited on

pages 75, 79, 81, and 122.)de Bruijn, N. and D. Klarner

1975. A finite basis theorem for pack-ing boxes with bricks. Philips Res.Rep, 30:337–343. (Cited on pages 185

and 224.)de Bruijn, N. G.

1969. Filling boxes with bricks.The American Mathematical Monthly,76(1):37–40. (Cited on page 157.)

Delest, M.-P. and J.-M. Fedou1993. Enumeration of skew Ferrersdiagrams. Discrete Mathematics, 112(1-3):65–79. (Cited on page 45.)

Delgado, O., D. Huson, and E. Zamorza-eva1992. The classification of 2-isohedraltilings of the plane. Geometriae Dedicata,42(1):43–117. (Cited on page 250.)

Desreux, S., M. Matamala, I. Rapaport,and E. Rémila2004. Domino tilings and related mod-els: space of configurations of domainswith holes. Theoretical computer science,319(1-3):83–101. (Cited on page 123.)

Diestel, R.2000. Graph Theory, volume 173 of Grad-uate Texts in Mathematics., 2nd edition.Springer-Verlag New York. (Cited onpage 68.)

Dietert, G.2010. An Integer linear programmingformulation for tiling large rectanglesusing 4 x 6 and 5 x 7 tiles. (Cited onpage 223.)

Doeraene, A.. Aztec Diamond Generator.https://sites.uclouvain.be/aztecdiamond/.(Cited on page 103.)

Donaldson, M.1996. Tiling with Dominoes. PhD thesis,Reed College. (Cited on page 69.)

Dubeau, F. and S. Labbé2007. EulerâAZs characteristics andPickâAZs theorem. International Journalof Contemporary Mathematical Sciences,2:909–928. (Cited on page 24.)

Easterfield, T. E.1946. A combinatorial algorithm. Jour-nal of the London Mathematical Society,1(3):219–226. (Cited on page 40.)

Elkies, N., G. Kuperberg, M. Larsen, andJ. Propp1992a. Alternating-sign matrices anddomino tilings (Part I). Journal of Alge-braic Combinatorics, 1(2):111–132. (Citedon page 102.)

Elkies, N., G. Kuperberg, M. Larsen, andJ. Propp1992b. Alternating-sign matrices and

https://math.stackexchange.com/q/570546

http://nms.lu.lv/wp-content/uploads/2015/11/Pentamino1.pdf




http://www.eklhad.net/polyomino/index.html

https://sites.uclouvain.be/aztecdiamond/

BIBLIOGRAPHY 259

domino tilings (Part II). Journal of Alge-braic Combinatorics, 1(3):219–234. (Citedon page 102.)

Engel, A.1998. Coloring Proofs. Problem-Solving Strategies, Pp. 25–37. (Cited onpages 27 and 68.)

Erickson, A.2013. Monomino-Domino Tatami Cover-ings. PhD thesis. (Cited on page 123.)

Erickson, A., F. Ruskey, J. Woodcock, andM. Schurch2011. Monomer-dimer tatami tilingsof rectangular regions. The ElectronicJournal of Combinatorics, 18(1):P109.(Cited on page 123.)

Erickson, A. and M. Schurch2012. Monomer-dimer tatami tilingsof square regions. Journal of DiscreteAlgorithms, 16:258–269. (Cited onpage 123.)

Eu, S.-P. and T.-S. Fu2005. A simple proof of the Aztecdiamond theorem. the electronic journalof combinatorics, 12(1):18. (Cited onpage 102.)

Fendler, M. and D. Grieser2016. A new simple proof of the Aztecdiamond theorem. Graphs and Com-binatorics, 32(4):1389–1395. (Cited onpage 102.)

Fletcher, R.1996. Tiling rectangles with symmetrichexagonal polyominoes. CongressusNumerantium, Pp. 3–29. (Cited onpages 173 and 174.)

Fogel, J., M. Goldenberg, and A. Liu2001. Packing Rectangles with Y-Pentominoes. Mathematics and Informat-ics Quarterly, 11(3):133–137. (Cited onpage 194.)

Fontaine, A.1991. An infinite number of planefigures with heesch number two. Jour-nal of Combinatorial Theory, Series A,57(1):151–156. (Cited on page 242.)

Fournier, J.-C.1996. Pavage des figures planes sanstrous par des dominos: Fondementgraphique de l’algorithme de Thurston,parallélisation, unicité et décompo-sition. Theoretical computer science,159(1):105–128. (Cited on page 35.)

Fournier, J.-C.1997. Tiling pictures of the planewith dominoes. Discrete mathematics,165:313–320. (Cited on page 68.)

Fricke, J.1995. Quadratzerlegung einesRechtecks. Mathematische Semester-

berichte, 42(1):53–62. (Cited onpages 158 and 159.)

Friedman, E.2008. Problem of the Month (January2008). Math Magic. https://www2.stetson.edu/~efriedma/mathmagic/0108.html.(Cited on page 15.)

Gershon, A.2015. New Directions in the Enumera-tion of Tilings on a Chessboard. (Citedon page 123.)

Gilhooley, C.2019a. If γ : [a, b] → C is contin-uous and γ(b) = −γ(a), must thecurves γ and eicγ intersect for all realc? Mathematics Stack Exchange.https://math.stackexchange.com/q/3114836(version: 2019-02-16). (Cited onpage 24.)

Gilhooley, C.2019b. Must a curve η : [a, b] → R2

intersect the curves η + η(b)−η(a)n

(n > 1)? Mathematics Stack Exchange.https://math.stackexchange.com/q/3125569(version: 2019-02-25). (Cited onpage 24.)

Golomb, S. W.1966. Tiling with polyominoes. Journalof Combinatorial Theory, 1(2):280–296.(Cited on pages 32, 124, 125, 126, 173,and 184.)

Golomb, S. W.1970. Tiling with sets of polyomi-noes. Journal of Combinatorial Theory,9(1):60–71. (Cited on page 244.)

Golomb, S. W.1981. Normed Division Domains.The American Mathematical Monthly,88(9):680–686. (Cited on page 252.)

Golomb, S. W.1996. Polyominoes: puzzles, patterns, prob-lems, and packings. Princeton UniversityPress. (Cited on pages 14, 23, 27, 32,50, 111, 112, 113, 114, 127, 175, 176, 184,253, and 254.)

Golomb, S. W. and D. A. Klarner1963. E1543. Covering a Rectangle withL-tetrominoes. The American Mathemat-ical Monthly, 70(7):760–761. (Cited onpage 190.)

Goodman-Strauss, C.2016. Tessellations. arXiv preprintarXiv:1606.04459. (Cited on page 249.)

Goupil, A., M.-E. Pellerin, and J. d. W.d’Oplinter2013. Partially directed snake polyomi-noes. arXiv preprint arXiv:1307.8432.(Cited on page 57.)

Graham, R. L.1981. Fault-free tilings of rectangles. In

The Mathematical Gardner, Pp. 120–126.Springer. (Cited on page 202.)

Gravier, S., J. Moncel, and C. Payan2007. A generalization of the pen-tomino exclusion problem: Dislocationof graphs. Discrete mathematics, 307(3-5):435–444. (Cited on page 253.)

Gravier, S. and C. Payan2001. On the pentomino exclusionproblem. Discrete & ComputationalGeometry, 26(3):375–386. (Cited onpage 253.)

Grekov, D. O.. Polyomino Tilings. http://polyominoes.org/. (Cited on pages 24,175, 184, 188, and 224.)

Grünbaum, B. and G. C. Shephard1977. The eighty-one types of isohedraltilings in the plane. In MathematicalProceedings of the Cambridge Philosoph-ical Society, volume 82, Pp. 177–196.Cambridge University Press. (Cited onpage 228.)

Grünbaum, B. and G. C. Shephard1987. Tilings and patterns. Freeman.(Cited on pages 24, 225, 226, 228,and 249.)

Grünbaum, B. and G. C. Shephard1993. Pick’s Theorem. The AmericanMathematical Monthly, 100(2):150–161.(Cited on page 24.)

Guttmann, A. J.2009. Polygons, polyominoes and poly-cubes, volume 775. Springer. (Cited onpage 23.)

Hall, P.1935. On representatives of subsets.Journal of the London Mathematical Soci-ety, 1(1):26–30. (Cited on page 39.)

Halmos, P. R. and H. E. Vaughan2009. The marriage problem. In ClassicPapers in Combinatorics, Pp. 146–147.Springer. (Cited on page 40.)

Harris, J. M., J. L. Hirst, and M. J. Moss-inghoff2008. Combinatorics and graph theory,volume 2. Springer. (Cited on page 68.)

Heesch, H. and O. Kienzle2013. Flächenschluss: System der For-men lückenlos aneinanderschliessenderFlachteile, volume 6. Springer-Verlag.(Cited on page 228.)

Herzog, J., S. S. Madani, et al.2014. The coordinate ring of a simplepolyomino. Illinois Journal of Mathemat-ics, 58(4):981–995. (Cited on page 19.)

Herzog, J., A. A. Qureshi, and A. Shikama2015. Gröbner bases of balanced poly-ominoes. Mathematische Nachrichten,288(7):775–783. (Cited on pages 13

https://www2.stetson.edu/~efriedma/mathmagic/0108.html

https://www2.stetson.edu/~efriedma/mathmagic/0108.html



http://polyominoes.org/

http://polyominoes.org/

260 polyominoes

and 32.)Hickerson, D.

2002. Filling rectangular rooms withtatami mats. https://oeis.org/A068920/a068920.txt. (Cited on page 111.)

Hitchman, M. P.2017. Tile Invariants for Tackling TilingQuestions. In A Primer for Undergradu-ate Research, Pp. 61–84. Springer. (Citedon page 68.)

Hochberg, R.2015. The gap number of the T-tetromino. Discrete Mathematics,338(1):130–138. (Cited on pages 42,106, 130, and 224.)

Honsberger, R.1973. Mathematical Gems, Math. Assoc,of America, P. 106. (Cited on page 57.)

Horne, C. E.2000. Geometric symmetry in patternsand tilings, volume 11. WoodheadPublishing. (Cited on page 249.)

Huson, D. H.1993. The generation and classifica-tion of tile-k-transitive tilings of theeuclidean plane, the sphere and thehyperbolic plane. Geometriae Dedicata,47(3):269–296. (Cited on pages 227

and 250.)Impellizieri, F.

2016. Domino Tilings of the Torus.arXiv preprint arXiv:1601.06354. (Citedon page 122.)

Ito, K.1996. Domino tilings on planar regions.journal of combinatorial theory, SeriesA, 75(2):173–186. (Cited on pages 71

and 122.)Jeandel, E. and M. Rao

2015. An aperiodic set of 11 Wang tiles.arXiv preprint arXiv:1506.06492. (Citedon page 244.)

Jepsen, C. H., L. Vaughn, and D. Brantley2003. Orders of L-shaped polyomi-noes. Journal of Recreational Mathematics,32(3):226. (Cited on page 174.)

JimmyK4542

2017. Do we need 17 monominoesto tile a 9 × 9 with square tetro-minoes? Mathematics Stack Ex-change. https://math.stackexchange.com/q/2577520(version:2017-12-23). (Cited onpage 164.)

Jockusch, W., J. Propp, and P. Shor1998. Random domino tilings and thearctic circle theorem. arXiv preprintmath/9801068. (Cited on page 103.)

Kaplan, C. S.2009. Introductory tiling theory forcomputer graphics. Synthesis Lec-

tures on Computer Graphics and Anima-tion, 4(1):35–54. (Cited on pages 226

and 249.)Kaplan, C. S.

2017. Heesch Numbers, Part2: Polyforms. http://isohedral.ca/heesch-numbers-part-2-polyforms/. (Citedon pages 7, 242, 243, and 244.)

Kass, S.2014. count the ways to fill a4 × n board with dominoes.Mathematics Stack Exchange.https://math.stackexchange.com/q/664236(version: 2014-02-06). (Cited onpage 98.)

Kasteleyn, P.1961. The statistics of dimers on alattice: I. The number of dimer arrange-ments on a quadratic lattice. Physica,27(12):1209 – 1225. (Cited on page 100.)

Kenyon, A. and M. Tassy2015. Tiling with Small Tiles. arXivpreprint arXiv:1511.03043. (Cited onpage 255.)

Kenyon, C.. Tiling groups and algorithms fortiling. (Cited on page 122.)

Kenyon, C. and R. Kenyon1992. Tiling a polygon with rectangles.IEEE. (Cited on page 122.)

Kenyon, C. and E. Rémila1996. Perfect matchings in the tri-angular lattice. Discrete Mathematics,152(1-3):191–210. (Cited on page 122.)

Kenyon, R.1993. Tiling with squares and square-tileable surfaces. Citeseer. (Cited onpage 122.)

Kenyon, R.1996. A note on tiling with integer-sided rectangles. journal of combinatorialtheory, Series A, 74(2):321–332. (Cited onpage 224.)

Kenyon, R.2000a. The asymptotic determinantof the discrete Laplacian. Acta Math-ematica, 185(2):239–286. (Cited onpage 123.)

Kenyon, R.2000b. Conformal invariance of dominotiling. Annals of probability, Pp. 759–795.(Cited on pages 37, 115, 118, and 123.)

Kenyon, R.2000c. The planar dimer model withboundary: a survey. Directions in math-ematical quasicrystals, CRM Monogr. Ser,13:307–328. (Cited on pages 35 and 68.)

Kenyon, R.2001. Dominos and the Gaussian freefield. Annals of probability, Pp. 1128–

1137. (Cited on page 123.)Kenyon, R. W., J. G. Propp, and D. B.

Wilson2001. Trees and matchings. Journal ofcombinatorics, 7(1):R25–R25. (Cited onpage 123.)

Klamkin, M. S. and A. Liu1980. Polyominoes on the infinitecheckerboard. Journal of CombinatorialTheory, Series A, 28(1):7–16. (Cited onpages 7, 253, and 254.)

Klarner, D. A.1965. Some results concerning poly-ominoes. (Cited on pages 165, 167,and 168.)

Klarner, D. A.1969. Packing a rectangle with con-gruent n-ominoes. Journal of Combina-torial Theory, 7(2):107–115. (Cited onpages 155, 156, 166, 173, and 198.)

Klarner, D. A.1973. A finite basis theorem revisited.Computer Science Department, Stan-dard University. (Cited on page 224.)

Klarner, D. A.1981. My life among the polyominoes.In The Mathematical Gardner, Pp. 243–262. Springer. (Cited on page 185.)

Klosinski, L. F., G. L. Alexanderson, andL. C. Larson1992. The fifty-second William LowellPutnam mathematical competition.The American Mathematical Monthly,99(8):715–724. (Cited on page 223.)

Knuth, D. E.2009. The Art of Computer Program-ming: Bitwise Tricks & Techniques;Binary Decision Diagrams, volume 4,fascicle 1. (Cited on page 111.)

Knuth, D. E., R. L. Graham, O. Patashnik,et al.1989. Concrete mathematics. AdisonWesley. (Cited on pages 97 and 123.)

Kolountzakis, M. N.2004. Filling a box with translatesof two bricks. the electronic journalof combinatorics, 11(1):16. (Cited onpage 158.)

Korn, M. and I. Pak2004. Tilings of rectangles with T-tetrominoes. Theoretical computer science,319(1-3):3–27. (Cited on page 224.)

Korn, M. R.2004. Geometric and algebraic proper-ties of polyomino tilings. PhD thesis,Massachusetts Institute of Technology.(Cited on page 49.)

Kranakis, E.1996. Characterization of DominoTilings of Squares with Prescribed

https://oeis.org/A068920/a068920.txt

https://oeis.org/A068920/a068920.txt

https://math.stackexchange.com/q/2577520 (version: 2017-12-23)

https://math.stackexchange.com/q/2577520 (version: 2017-12-23)

http://isohedral.ca/heesch-numbers-part-2-polyforms/

http://isohedral.ca/heesch-numbers-part-2-polyforms/


BIBLIOGRAPHY 261

Number of Nonoverlapping 2 2

Squares. (Cited on pages 87, 88, 90,92, and 94.)

Krattenthaler, C.2000. Schur function identities andthe number of perfect matchings ofholey Aztec rectangles. In Q-series froma Contemporary Perspective: AMS-IMS-SIAM Joint Summer Research Conferenceon Q-Series, Combinatorics, and ComputerAlgebra, June 21-25, 1998, Mount HolyokeCollege, South Hadley, MA, volume 254,P. 335. American Mathematical Soc.(Cited on page 123.)

Kung, J. P., G.-C. Rota, and C. H. Yan2009. Combinatorics: the Rota way. Cam-bridge University Press. (Cited onpage 40.)

Labrousse, D. and J. Ramírez Alfonsín2010. A Tiling Problem and the Frobe-nius Number. In Additive NumberTheory, Pp. 203–220. Springer. (Citedon pages 159, 160, and 162.)

Lai, T.2013. New aspects of regions whosetilings are enumerated by perfect pow-ers. arXiv preprint arXiv:1309.6022.(Cited on page 123.)

Lai, T.2014. A Simple Proof for the Number ofTilings of Quartered Aztec Diamonds.Electr. J. Comb., 21:P1.6. (Cited onpage 123.)

Lam, F. and L. Pachter2003. Forcing numbers of stop signs.Theoretical computer science, 303(2-3):409–416. (Cited on page 68.)

Langerman, S. and A. Winslow2015. A quasilinear-time algorithm fortiling the plane isohedrally with a poly-omino. arXiv preprint arXiv:1507.02762.(Cited on page 228.)

Leroux, P., E. Rassart, and A. Robitaille1998. Enumeration of symmetry classesof convex polyominoes in the squarelattice. arXiv preprint math/9803130.(Cited on page 45.)

Liu, A.2018a. A Sample Minicourse: Tessel-lations, Pp. 27–60. Cham: SpringerInternational Publishing. (Cited onpage 124.)

Liu, A. C.-F.2018b. Polyform Compatibility, Pp. 39–53. Cham: Springer InternationalPublishing. (Cited on page 252.)

Lodi, A., S. Martello, and M. Monaci2002. Two-dimensional packing prob-lems: A survey. European journal ofoperational research, 141(2):241–252.

(Cited on page 223.)Los, J., J. Mycielski, D. C. B. Marsh, and

M. Goldberg1960. E1390. Primitive Decompositions.The American Mathematical Monthly,67(5):477–478. (Cited on page 220.)

Lovász, L. and M. D. Plummer2009. Matching theory, volume 367.American Mathematical Soc. (Cited onpage 68.)

MacMahon, P. A.1921. New mathematical pastimes. TheUniversity Press. (Cited on page 244.)

Maltby, S. J.1994. Trisecting a rectangle. Jour-nal of Combinatorial Theory, Series A,66(1):40–52. (Cited on page 223.)

Mann, C.2004. Heesch’s tiling problem. TheAmerican Mathematical Monthly,111(6):509–517. (Cited on page 242.)

Mann, C. and B. C. Thomas2016. Heesch numbers of edge-markedpolyforms. Experimental Mathematics,25(3):281–294. (Cited on page 242.)

Marshall, W. R.1997. Packing rectangles with congru-ent polyominoes. Journal of Combina-torial Theory, Series A, 77(2):181–192.(Cited on pages 174, 176, 184, and 185.)

Martin, G.1991. Polyominoes: A guide to puzzles andproblems in tiling. Cambridge UniversityPress. (Cited on pages 23, 27, 112, 156,190, 218, 219, and 237.)

Martin, G. E.1986. Polytaxic Polygons. Struc-tural Topology 1986 múm 12. (Cited onpages 45 and 238.)

Mason, J.2014. Chessboard coloured polyomi-noes and bias. (Cited on pages 32

and 33.)Mathar, R. J.

2013. Paving rectangular regionswith rectangular tiles: Tatami andnon-tatami tilings. arXiv preprintarXiv:1311.6135. (Cited on page 123.)

Mathar, R. J.2014. Tilings of rectangular regionsby rectangular tiles: Counts derivedfrom transfer matrices. arXiv preprintarXiv:1406.7788. (Cited on page 123.)

Mendelsohn, N. S.2004. Tiling with Dominoes. The Col-lege Mathematics Journal, 35(2):115–120.http://www.jstor.org/stable/4146865. (Citedon pages 27, 31, and 68.)

Merino, C.2008. On the number of tilings of the

rectangular board with T-tetronimoes.AUSTRALASIAN JOURNAL OFCOMBINATORICS, 41:107. (Citedon page 224.)

Milet, P. H.2015. Domino tilings of three-dimensional regions. arXiv preprintarXiv:1503.04617. (Cited on page 122.)

Muchnik, R. and I. Pak1999. On tilings by ribbon tetrominoes.Journal of Combinatorial Theory Series A,88:188–193. (Cited on page 122.)

Myers, J.. Polyomino, polyhex and polyiamondtiling. https://www.polyomino.org.uk/mathematics/polyform-tiling/. (Cited onpages 7, 24, 228, 229, 230, 231, 234, 235,236, 238, and 240.)

Narayan, D. A. and A. J. Schwenk2002. Tiling large rectangles. Mathemat-ics magazine, 75(5):372–380. (Cited onpage 223.)

Neretin, I.2017. Does every domino tilinghave at least two “exposed” domi-noes? Mathematics Stack Exchange.https://math.stackexchange.com/q/2467370(version: 2017-10-11). (Cited onpage 46.)

nickgard2017. Are tilings by completelysymmetric polyominoes alwaysunique? Mathematics Stack Exchange.https://math.stackexchange.com/q/2543241(version: 2017-11-29) (Author URL:https://math.stackexchange.com/users/420432/nickgard). (Cited on page 15.)

Nitica, V.2004. Tiling a deficient rectangle byL-tetrominoes. Journal of RecreationalMathematics, 33(4):259. (Cited onpage 224.)

Nitica, V.2018a. Revisiting a Tiling Hierarchy.IEEE Transactions on Information Theory,64(4):3162–3169. (Cited on page 132.)

Nitica, V.2018b. Revisiting a Tiling Hierarchy(II). Open Journal of Discrete Mathematics,8(02):48. (Cited on page 132.)

Niven, I. and H. Zuckerman1967. Lattice points and polygonal area.The American Mathematical Monthly,74(10):1195–1200. (Cited on page 20.)

Oh, S.2018. Domino tilings of the expandedAztec diamond. Discrete Mathematics,341(4):1185–1191. (Cited on page 123.)

Oliveira e Silva, T.2015. Animal enumerations

http://www.jstor.org/stable/4146865

https://www.polyomino.org.uk/mathematics/polyform-tiling/

https://www.polyomino.org.uk/mathematics/polyform-tiling/



https://math.stackexchange.com/users/420432/nickgard

https://math.stackexchange.com/users/420432/nickgard

262 polyominoes

on the {4,4} Euclidean tiling.http://sweet.ua.pt/tos/animals/a44.html.(Cited on page 24.)

O’rourke, J.1987. Art gallery theorems and algorithms,volume 57. Oxford University PressOxford. (Cited on page 24.)

Pachter, L.1997. Combinatorial approaches andconjectures for 2-divisibility prob-lems concerning domino tilings ofpolyominoes. The Electronic Journalof Combinatorics, 4(1):29. (Cited onpage 123.)

Pachter, L. and P. Kim1998. Forcing matchings on squaregrids. Discrete Mathematics, 190(1-3):287–294. (Cited on pages 59 and 68.)

Pak, I.2000. Ribbon tile invariants. Trans-actions of the American MathematicalSociety, 352(12):5525–5561. (Cited onpage 45.)

Parlier, H. and S. Zappa2017. Distances in domino flip graphs.American Mathematical Monthly,124(8):710–722. (Cited on pages 56,83, and 84.)

Pegg, E., A. H. Schoen, and T. Rodgers2009. Homage to a pied puzzler. AKPeters Ltd. (Cited on page 252.)

Penrose, R.1994. Shadows of the Mind, volume 4.Oxford University Press Oxford. (Citedon page 245.)

Pick, G.1899. Geometrisches zur zahlenlehre.Sitzenber. Lotos (Prague), 19:311–319.(Cited on page 20.)

Plummer, M. D.1992. Matching theory–a sampler: fromDénes König to the present. DiscreteMathematics, 100(1-3):177–219. (Cited onpage 68.)

Preparata, F. P. and M. I. Shamos2012. Computational geometry: an in-troduction, chapter 8, Pp. 323–373.Springer Science & Business Media.(Cited on page 24.)

Propp, J.1997. A pedestrian approach to amethod of Conway, or, A tale of twocities. Mathematics Magazine, 70(5):327–340. (Cited on page 68.)

Propp, J.2002. Lattice structure for orientationsof graphs. arXiv preprint math/0209005.(Cited on page 62.)

Propp, J. et al.1999. Enumeration of matchings: prob-

lems and progress. New Perspectives inGeometric Combinatorics (eds. L. Billera,A. Björner, C. Greene, R. Simeon, and RPStanley), Cambridge University Press,Cambridge, Pp. 255–291. (Cited onpage 123.)

Propp, J. and U. M. Lowell2015. Enumeration of tilings. Handbookof Enumerative Combinatorics. CRC Press,Boca Raton, 84. (Cited on page 123.)

Propp, J. and R. Stanley1999. Domino tilings with barriers.Journal of Combinatorial Theory, Series A,87(2):347–356. (Cited on page 122.)

Puleo, G. J.2018a. How many colors are nec-essary for a rectangle to nevercover a color more than once?Mathematics Stack Exchange.https://math.stackexchange.com/q/3037153(version: 2018-12-13). (Cited onpage 135.)

Puleo, G. J.2018b. How many colors are nec-essary for a rectangle to nevercover a color more than once?Mathematics Stack Exchange.https://math.stackexchange.com/q/3047642(version: 2018-12-20). (Cited onpage 138.)

Ramírez-Alfonsín, J. L.2005. The Diophantine Frobenius problem,volume 30. Oxford University Press onDemand. (Cited on page 223.)

Redelmeier, D. H.1981. Counting polyominoes: yetanother attack. Discrete Mathematics,36(2):191–203. (Cited on page 14.)

Reid, M.. Michael Reid’s polyomino page.http://www.cflmath.com/~reid/Polyomino/.(Cited on pages 24, 174, 175, 182, 194,195, 196, and 197.)

Reid, M.1997. Tiling rectangles and half stripswith congruent polyominoes. Jour-nal of Combinatorial Theory, Series A,80(1):106–123. (Cited on pages 132, 174,and 182.)

Reid, M.1998. Tiling a square with eight congru-ent polyominoes. Journal of Combinato-rial Theory Series A, 83(1):158. (Cited onpage 184.)

Reid, M.2003. Tile homotopy groups. EN-SEIGNEMENT MATHEMATIQUE,49(1/2):123–156. (Cited on pages 126

and 197.)

Reid, M.2005. Klarner systems and tiling boxeswith polyominoes. J. Comb. Theory, Ser.A, 111(1):89–105. (Cited on pages 111,185, 194, 195, 197, and 224.)

Reid, M.2008. Asymptotically optimal boxpacking theorems. the electronic journalof combinatorics, 15(1):78. (Cited onpage 224.)

Reid, M.2014. Many L-Shaped PolyominoesHave Odd Rectangular Packings. An-nals of Combinatorics, 18(2). (Cited onpages 168, 173, 174, and 184.)

Rémila, E.1996. Tiling a figure using a height in atree. In SODA, Pp. 168–174. (Cited onpage 122.)

Rémila, E.2004. The lattice structure of the set ofdomino tilings of a polygon. Theoreticalcomputer science, 322(2):409–422. (Citedon pages 67, 69, 71, 72, 73, 74, 76, 80,and 122.)

Resta, G.. Polypolyominoes. http://www.iread.it/Poly/index.php (Accessed 10 May 2019).(Cited on page 252.)

Rhoads, G. C.2003. Planar Tilings and the Search foran Aperiodic Prototile. PhD thesis, NewBrunswick, NJ, USA. AAI3092982.(Cited on page 243.)

Rhoads, G. C.2005. Planar tilings by polyominoes,polyhexes, and polyiamonds. Journal ofComputational and Applied Mathematics,174(2):329–353. (Cited on page 226.)

Robinson, P. J.1982. Fault-free rectangles tiled withrectangular polyominoes. In Combina-torial Mathematics IX, E. J. Billington,S. Oates-Williams, and A. P. Street,eds., Pp. 372–377, Berlin, Heidelberg.Springer Berlin Heidelberg. (Cited onpages 202 and 205.)

Roman, S.2008. Lattices and ordered sets. SpringerScience & Business Media. (Cited onpage 122.)

Ruskey, F. and J. Woodcock2009. Counting fixed-height tatamitilings. The Electronic Journal of Combi-natorics, 16(1):126. (Cited on pages 111

and 123.)Saldanha, N. C. and C. Tomei

2003. Tilings of quadriculated annuli.Journal of Combinatorial Theory, Series B,88(1):153–183. (Cited on page 122.)

http://sweet.ua.pt/tos/animals/a44.html



http://www.cflmath.com/~reid/Polyomino/

http://www.iread.it/Poly/index.php

http://www.iread.it/Poly/index.php

BIBLIOGRAPHY 263

Saldanha, N. C., C. Tomei, M. Casarin, andD. Romualdo1995. Spaces of domino tilings. Discrete& Computational Geometry, 14(1):207–233. (Cited on pages 34, 62, 67, 69, 122,and 123.)

Sallows, L., M. Gardner, R. K. Guy, andD. Knuth1991. Serial isogons of 90 degrees.Mathematics Magazine, 64(5):315–324.(Cited on page 17.)

Sallows, L. C.1992. New pathways in serial iso-gons. The Mathematical Intelligencer,14(2):55–67. (Cited on page 17.)

Saxton Jr, M. A.2015. Tiling with Polyominoes, Polycubesand Rectangles. PhD thesis, University ofCentral Florida Orlando, Florida. (Citedon page 193.)

Sayranov, A.. Holes in double-tileable poly-nominoes. MathOverflow.https://mathoverflow.net/q/317284 (ver-sion: 2018-12-10). (Cited on page 24.)

Schattschneider, D. and M. C. Escher1990. Visions of symmetry: Notebooks,periodic drawings, and related work ofMC Escher. WH Freeman. (Cited onpage 228.)

Scherer, K.1980. Minimal fault-free rectan-gles packed with In-polyominoes. J.Recreational Math, 13:4–6. (Cited onpage 205.)

Scherphuis, J.2016. G4G12 Exchange Gift: Ziggy.https://www.jaapsch.net/g4g/g4g12.htm(Accessed 12 April 2019). (Cited onpage 140.)

Sicherman, G.. Polyform Compatibility. http://www.recmath.org/PolyCur/compatibility.html,https://userpages.monmouth.com/~colonel/compatibility.html (Accessed 10 May2019). (Cited on pages 251 and 252.)

Sicherman, G.2015. Baiocchi Figures for Polyominoes.http://www.recmath.org/PolyCur/nbaiocchi/.(Accessed on 04/06/2019). (Cited onpage 16.)

Sillke, T.1992. Prime Packings of the PentominoY. https://www.math.uni-bielefeld.de/~sillke/PENTA/qu5-y. (Cited on page 194.)

Simon, D. P.2016. Tilings and the Aztec Diamond The-orem. PhD thesis. (Cited on page 102.)

Sloane, N. J. A.. Encyclopedia of Online Integer Se-

quences. https://oeis.org/. (Cited onpage 10.)

Socolar, J. E. and J. M. Taylor2011. An aperiodic hexagonal tile.Journal of Combinatorial Theory, Series A,118(8):2207–2231. (Cited on page 226.)

Soifer, A.2010. Geometric etudes in combinato-rial mathematics. Springer Science &Business Media. (Cited on page 23.)

Stanley, R. P.1986. What Is Enumerative Combina-torics? In Enumerative combinatorics,Pp. 1–63. Springer. (Cited on page 82.)

Stewart, I. N. and A. Wormstein1992. Polyominoes of order 3 do not ex-ist. Journal of Combinatorial Theory, SeriesA, 61(1):130–136. (Cited on page 184.)

Stucky, E.2015. An Exposition of Kasteleyn’sSolution of the Dimer Model. (Cited onpage 100.)

Tauraso, R.2004. A new domino tiling sequence.Journal of Integer Sequences, 7(04.2):3.(Cited on page 123.)

Taxel, P.2016. Can a rectangle be tiled with 6smaller rectangles, such that no smallerset of those rectangles forms a rect-angle? Mathematics Stack Exchange.https://math.stackexchange.com/q/1895836(version: 2016-08-18). (Cited onpage 221.)

Taylor, P.2018. How many colors are nec-essary for a rectangle to nevercover a color more than once?Mathematics Stack Exchange.https://math.stackexchange.com/q/3043244(version: 2018-12-31). (Cited onpage 139.)

Thiant, N.2003. An O (nlogn)-algorithm for find-ing a domino tiling of a plane picturewhose number of holes is bounded.Theoretical Computer Science, 303(2-3):353–374. (Cited on pages 32, 45,and 123.)

Thurston, W. P.1990. Conway’s tiling groups. Amer.Math. Monthly, 97(8):757–773. (Cited onpages 69, 82, 85, and 122.)

Toth, C. D., J. O’Rourke, and J. E. Good-man2017. Handbook of discrete and com-putational geometry. Chapman andHall/CRC. (Cited on pages 13, 23,and 249.)

Tremblay, H.2019. Intersection of translatedsimple curves in R2. Mathe-matics Stack Exchange. https://math.stackexchange.com/q/1322791 (ver-sion: 2015-06-12). (Cited on page 24.)

Tulleken, H.2019. Will a path between (x, y)and (−x,−y) always intersect a90 degree rotated copy? Math-ematics Stack Exchange. https://math.stackexchange.com/q/3106208 (ver-sion: 2019-02-17). (Cited on page 24.)

Varberg, D. E.1985. Pick’s Theorem Revisited.The American Mathematical Monthly,92(8):584–587. (Cited on page 24.)

von Eitzen, H.2019. A continuous curve inter-sects its 90 degrees rotated copy?Mathematics Stack Exchange.https://math.stackexchange.com/q/3109394(version: 2019-02-12). (Cited onpage 23.)

Wagon, S.1987. Fourteen proofs of a result abouttiling a rectangle. The American Math-ematical Monthly, 94(7):601–617. (Citedon page 223.)

Walkup, D.1965. Covering a rectangle with T-tetrominoes. The American MathematicalMonthly, 72(9):986–988. (Cited onpages 190, 191, 192, and 193.)

Wang, H.1961. Proving theorems by patternrecognitionâATII. Bell system technicaljournal, 40(1):1–41. (Cited on page 244.)

Wikipedia contributors2017. Rectilinear polygon —Wikipedia, The Free Encyclopedia.https://en.wikipedia.org/w/index.php?title=Rectilinear_polygon&oldid=793031242.[Online; accessed 24-September-2018].(Cited on pages 16, 17, and 18.)

Wilf, H. S.2005. generatingfunctionology. AKPeters/CRC Press. (Cited on page 123.)

Williams, W. and C. Thompson2008. The Perimeter of a Polyominoand the Surface Area of a Polycube. TheCollege Mathematics Journal, 39(3):233–237. (Cited on pages 20 and 21.)

Winslow, A.2015. An Optimal Algorithm forTiling the Plane with a Translated Poly-omino. In ISAAC. (Cited on pages 226

and 247.)Winslow, A.

2018. Some Open Problems in Poly-

https://mathoverflow.net/q/317284

https://www.jaapsch.net/g4g/g4g12.htm

http://www.recmath.org/PolyCur/compatibility.html

http://www.recmath.org/PolyCur/compatibility.html

https://userpages.monmouth.com/~colonel/compatibility.html

https://userpages.monmouth.com/~colonel/compatibility.html

http://www.recmath.org/PolyCur/nbaiocchi/

https://www.math.uni-bielefeld.de/~sillke/PENTA/qu5-y

https://www.math.uni-bielefeld.de/~sillke/PENTA/qu5-y

https://oeis.org/








https://en.wikipedia.org/w/index.php?title=Rectilinear_polygon&oldid=793031242

https://en.wikipedia.org/w/index.php?title=Rectilinear_polygon&oldid=793031242

264 polyominoes

omino Tilings. In International Con-ference on Developments in LanguageTheory, Pp. 74–82. Springer. (Cited onpages 23, 124, 127, 182, 228, and 238.)

Wolfram, S.2002. A new kind of science, volume 5.Wolfram media Champaign, IL. (Citedon page 246.)

Yang, J.2014. Rectangular tileability andcomplementary tileability are undecid-able. European Journal of Combinatorics,41:20–34. (Cited on page 244.)

YiFan2019. A continuous curve inter-sects its 90 degrees rotated copy?Mathematics Stack Exchange.https://math.stackexchange.com/q/3109299(version: 2019-02-12). (Cited onpage 24.)

Yu, H.2014. Polyominoes on a MulticoloredInfinite Grid. WORLD FEDERATIONOF NATIONAL MATHEMATICS COM-PETITIONS, P. 58. (Cited on pages 133

and 134.)Yuan, L., C. T. Zamfirescu, and T. I. Zam-

firescu2016. Dissecting the square into fivecongruent parts. Discrete Mathematics,339(1):288–298. (Cited on page 224.)

Zhan, S.2012. Tiling a deficient rectangle witht-tetrominoes. preprint (August 2752012). (Cited on page 224.)

Zhang, H.1996. The connectivity of Z-transformation graphs of perfectmatchings of polyominoes. DiscreteMathematics, 158(1-3):257–272. (Cited onpage 68.)


Index

Page numbers in bold point to a definition, numbers in italics point to an image, table or example.

k-anisohedral, 228k-isohedral, 227m-morphic, 237n-omino, 11p-poic, 2374-polypillar, 242

A-hexomino, 134, 195, 196

A000045, 96

A000105, 10, 57

A000330, 84

A001835, 98

A002013, 57

A005178, 98

A006125, 102

A046092, 102

A056785, 121

A099390, 100

A131482, 34

A210996, 121

A213377, 121

A234012, 121

A234013, 33

absorbtion, 79

algebra of domino tilingsextra reading, 68

algorithmdomino tiling

by cylinder removal, 43

of even polyomino, 118

of polyomino with evensides, 38

regions with holes, 123

strip rotation by flipsclosed strip with holes,

66

closed strip withoutholes, 65

tiling by rectangles, 224

tiling from height func-tion, 72

anti-knob, 18anti-symmetric, 74

aperiodic, 226Arctic behavior, 101

Arctic Circle Theorem, 103area criterion, 30

implications, 30–31

associative, 79

average area, 223axis-aligned, 16Aztec diamond, 100, 101, 123

BN factorization, 227

classification diagram, 46

fault-free, 115

maximum rank, 84

back, 170bad, 40Baiocchi figure, 15balanced, 32

enumeration, 119

Barclassification diagram, 46

bar, 38, 122, 224

gap number, 163

bar graph, 45, 170


gap number, 107

barrel, 170barrier, 122

basic, 201basic rectangle, 173bias, 32biased, 32bibone, 122

black, 55black patch, 39block, 193blunt end, 170blunt gun, 170bond pattern, 111

border, 36

border cell, 11border crossing theorem, 30

brick, 223

bridge, 30

C-hexomino, 195

cells, 11centrosymmetric, 182, 227

centrosymmetric curve, 184

chain, 193, 224

charge, 157

checkerboard coloring, 32, 53,142

266 polyominoes

checkerboard criterion, 32

chromatic number, 132closed, 53closed strip


color argument, 68

coloring, 254

hull, 133

column convex polyominoclassification diagram, 46

gap number, 107

combinatoricsfurther reading, 68

common multiple, 252commutative, 74, 79

compact, 29, 59compact subregion, 59comparable, 47compatibility, 24, 251

compatible, 251concave, 17connected, 13contractions, 42convex, 17convex polyomino


core, 56corner, 111

corner cell, 11, 36cornerless, 190covering relation, 75coverings, 68crossed strip, 126

cut, 35, 190Cylinder


cylinder, 42, 61, 113, 115, 126

gap number, 107

D-hexomino, 171, 195, 197

de Bruijn section, 42deficiency, 32deficient Aztec rectangle, 123

degenerate gun, 170

deletion, 42dilation, 57dimer, 68dimer model, 68, 123

direct path, 85discriminating coloring, 51distance, 85distributive lattice, 70, 81domino, 11, 225

domino edge path, 71domino tiling

enumeration, 96

down, 73

E-hexomino, 195

edgecolored, 36

efficient, 133efficient polyomino, 253

end, 170end block, 112

enumerationbalanced polyominoes,

119

domino tilings, 96

polyominoes tileable bydominoes, 119

tilings, 123

tilings of rectangle withdominoes and onemonomino, 109

erosion, 57even, 165even polyomino, 115even side, 37, 115

exact, 226exact polyomino, 134, 226

exclusion, 253

expanded Aztec diamond, 123

exposed, 46exposed long edges, 46

extension, 41, 113

extensions, 201

F-hexomino, 195

F-pentomino, 193

face, 170face twist, 62factorization theorem, 123

families, 184fault, 111, 201fault-free, 111Ferrers diagram, 45Fibbonacci number, 96

figure, see region, 11filled polyomino, 38

fixed, 14fixed single edge, 68flag coloring, 50, 132flat, 18, 115

flip, 62, 70

flippable pair, 62, 64–67, 70,73, 75, 90–92, 218, 225

flow, 34flow theorem, 34

forcing number, 68fountain set, 255Fourier transform, 158

fracture edges, 68free, 14Frobenius number, 159, 223

frozen, 30, 53, 58fundamental domain, 226fundamental region, 226

G-hexomino, 195, 197

gap, 170gap number, 106, 163

bricks by rods, 223

even polyominoes, 119

large rectangles by rect-angles, 223

rectangles by harmonicbricks, 223

golygon, 17, 227

good, 40graph, 123, 253

gun, 170Gun Theorem, 170

INDEX 267

H-hexomino, 195

half cylinder, 209half-strip, 225

harmonic brick, 223

Hasse diagram, 76

Heesch number, 242Heesch number 2, 243, 244

height difference, 71height function, 71

region with holes, 122

heptomino, 25

herringbone pattern, 111

herringbone pattern., 114

hexomino, 25

tileable by dominoes, 121

high 4-hexomino, 195

High F-hexomino, 195

high Y-hexomino, 195

hole, 13, 19, 116, 241

relation to mountains andvalleys, 19

holey, 13holy square, 123

horizontal n-cylinder, 42horizontal bond pattern, 111

hull, 125–127, 133, 167, 167–170, 193

I-hexomino, 195

idempotent, 79

inefficient, 133interior cell, 11interior vertex, 20, 85

intersection free polyominoclassification diagram, 46

isohedral, 227isohedral-tiling

kind, 228

isolated, 51italic X-hexomino, 195

J-hexomino, 195, 196

jig-saw, 49join, 78

K-hexomino, 195

knob, 18

L-configuation, 88L-hexomino, 195

L-shape, 222

L-shaped domain, 123

L-shaped polyomino, 107, 226


l-shaped-polyomino, 173

L-tetromino, 190

L-tetrominoes, 224

lattice, 79lattice polygon, 20leaning domino, 122

lev, 56

level, 56line of cleavage, 111local maximum, 85local minimum, 85long N-hexomino, 195

long S-hexomino, 195

long T-hexomino, 195

low 4-hexomino, 195

Low F-hexomino, 195

low Y-hexomino, 195

lozenge, 122

LR-configuration, 88

M-hexomino, 195

MacMahon tile, 244Manhatten polyomino, 45marriage theorem, 39

further reading, 68

matching theory, 68matchings

enumeration, 123

maximally biased, 33maximum perimeter, 34

maximum tiling, 81meet, 79mid-block, 112

minimal, 16minimum perimeter, 22

minimum tiling, 81, 123

monohedral, 225monomer, 68monomino, 11, 115, 253

moon polyominoclassification diagram, 46

mountain, 18, 115

mutilated chessboard prob-lem, 27, 31

neighbors, 11non-compact, 29

non-periodic, 225non-tiler, 241notation, 10

number of levels, 56

O-hexomino, 195

odd, 165odd rectangular order, 165odd side, 37, 115

one-sided, 14open problem, 16, 17, 24, 124,

172, 182, 185, 226,232, 249, 251

optimal, 106optimal coloring, 132, 253

order, 153order-2 polyominoes, 182

order-3 polyominoes, 184

order-4 polyominoes, 184

orthogonal polygon, 16

P pentomino, 127

P-hexomino, 195

P-octomino, 198

parityhorizontal and vertical

dominoes, 31

partial order, 74

partition, 13partition polyomino, 45partitions, 30

path, 53patriotic coloring, 50peak, 56, 59

268 polyominoes

pentomino, 11exclusion, 254

perfect, 223perimeter criterion, 22

Pick’s Theorem, 20, 21, 24, 232

picture, see region, 11planar dimer tilings, 123plane, 111, 226

polydomino, 119polygon, 17

polyomino, 11defined by borders, 116,

117

prime, 129prime rectangle, 185, 201, 224

prime rectangles, 153problems

domino, 122

open, 24, 185, 249

pentomino, 24

profane, 13proper patch, 39proper subregion, 13protects, 133puzzle, 140

pyramid, 87Pyramid Lemma, 88

Q-hexomino, 195

quadrant, 126, 127, 190

quadriculated surface, 122

quartered Aztec diamond, 123

R-configuration, 88R-hexomino, 195

rank, 82rectangle, 56, 107, 118, 125,

126, 226

arctic behaviour, 103



closed strip (is a), 55

coloring, 135, 138

counting tilings, 96

deficient, 108

dissection, 224

exposure, 47

fault-free, 111

gap number, 107, 164

maximum rank, 84

minimum and maximumtilings, 81

notation, 96

packing, 164

Saturnian (is), 56

tileability, 54

tiling with, 122

rectangular order, 165rectifiable, 165rectifiable polyomino, 251

rectilinear polygon, 16, 24

recurrence relation, 123

reflexive, 74

region, 11replication, 125, 126

reptile, 124, 125, 167reverse, 151right tromino, 153

ring, 53, 57classification diagram, 46

Robert I. Jewett, 111

rod, 223

row convex, 48row convex polyomino


gap number, 107

row-criterion, 22

S-pentomino, 193

safe, 43saturnian, 56serial isogon, 17short N-hexomino, 195

short T-hexomino, 195

short Z-hexomino, 195

simple, 13, 218simply-connected, 13, 165

simply-connected plyominoclassification diagram, 46

simply-connected polyomino,19

smallest common multiple,252

snake, 57classification diagram, 46

gap number, 107

solid gun, 170spanning tree, 123

spin, 71square, 92, 122, 125, 162, 201

gap number, 164

packing, 164

square coloring, 132, 164

square order, 165square tetromino, 18, 22

square-tetromino, 116

stack polyomino, 48, 67, 87


staircase, 45start block, 112

strip, 225, 249

strip polyomino, 53classification diagram, 46

gap number, 107

strip rotation, 62, 70

strip shift, 110strong prime rectangle, 185subregion, 13substitution tiling, 125sufficient coloring, 132symmetry, 14, 91, 123, 152,

182, 184

symmetry index, 14

T-pentomino, 131, 193

T-tetromino, 33, 125, 171, 190,192, 224

tab, 170tall Z-hexomino, 195

tatami tiling, 111, 114, 123

Temperleyan polyomino, 118,123

tetromino, 11, 122

INDEX 269

border word, 151

border words, 151

tileable by dominoes, 119

thickness, 170three-dimensional region, 122

tile-k-transitive, 227tile-transitive, 227tiles, 11tiling, 12, 24

tiling criteriadomino tilings (sum-

mary), 120

tiling invariant, 68

tiling problems, 11tilings

enumeration, 123

torus, 122

tower, 112

transitively equivalent, 227transitivity classes, 227trapeze, 48trapezoidal polyomino, 45triangle, 122

tribone, 122

tromino, 11two-color criterion, 51type-A vertex, 191type-B vertex, 191

U-frame polyomino, 242U-hexomino, 195

U-pentomino, 193

unbalanced polyomino, 29

unbiased, 32unfair, 51uniform sampling, 123

universal block, 112

up, 73

V-hexomino, 195

V-pentomino, 131, 193

valley, 18, 115

value, 71vertical n-cylinder, 42vertical bond pattern, 111

W-pentomino, 193

W-polyomino, 140BN factprization, 227

W-polyominoes, 133

Wa-hexomino, 195

Wang tile, 244water strider, 123

Wb-hexomino, 195

Wc-hexomino, 195

weak, 133white, 55white patch, 39word, 151

X-hexomino, 195

X-pentomino, 193

Y pentomino, 127

Y-hexomino, 171, 197

Y-pentomino, 153, 171

Y-polyominoBN factorization, 227

Young diagram, 45, 226


gap number, 107

Z-pentomino, 193

zig-zag polyominoes, 140

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